Chapter 4. Two-Dimensional Motion

09/12/2003

Chapter 4. Two-Dimensional Motion

I. Intuitive (Understanding) Review Problems.

If a car (object, body, truck) moves with positive velocity and negative acceleration, it means that its…

a)speed is decreasing;

b)speed is increasing;

c)acceleration is increasing;

d)acceleration is decreasing;

e)speed is constant.

Comment:

Whenever the velocity and acceleration vectors point in opposite directions, the speed of the object decreases.

If a car moves with constant velocity, it means that its…

a)acceleration is increasing;

b)acceleration is decreasing;

c)speed is decreasing;

d)speed is increasing;

e)speed is constant.

Comment:

If an object’s velocity is constant, the speed must also be constant, which implies that the acceleration is zero.

If the acceleration of an object is zero, what can you say about its velocity?

a)velocity is zero;

b)velocity is positive;

c)velocity is negative;

d)velocity is constant;

e)velocity is increasing;

f)velocity is decreasing;

g)question cannot be answered without additional information.

Comment:

Acceleration is defined as the change in velocity per change in time. Thus, if the acceleration is zero, the velocity is not changing.

An object (car, truck, body) is moving with constant acceleration of 2 m/s2. What is its acceleration 3 s later?

Comment:

The acceleration is constant, thus it is not changing and will be the same any time later.

At t = 0 s, a car (body, truck, object) starts to accelerate from rest at 3 m/s2. What is the speed of the car at the end of 4 s? (If the acceleration remains constant, what is the change in velocity of the car over the first 4 s? What is the distance the car travels during the first 4 s?)

Comment:

To find the speed (or velocity) of the car, use the equation v = v0 + at, where v0 = 0 m/s (since the object starts moving from rest).

Change in velocity is simply the final velocity minus the initial velocity.

To find the distance the object traveled, use the equation x = v0t + (½)at2.

A car (body, truck, object) decelerates from 65 mi/h to 40 mi/h. Find the magnitude of deceleration of the car if the stopping distance is equal to 300 m.

Comment:

To find the deceleration of the car, use the equation v2 = v02 + 2ax.

II. Projectile Motion: Zero Launch Angle

A tennis ball is thrown horizontally with the initial speed of 31 m/s. How much time is the ball in the air if it lands 155 m away?

Hint 1:

Projectile motion problems can be solved by analyzing the motion in the x and y directions separately.

The acceleration in the x direction is zero. Because the acceleration in the y direction is constant, the equations for constant acceleration apply. Time is the variable that links the x and y equations together.

Remember that the motions along x and y axis are completely independent of each other.

Comment:

The data provided in the text of the problem describes only the motion in the x direction, so in order to find how long the ball is in the air, use the equation x = v0xt + (½)axt2. Note that the velocity in the x direction is constant, so the acceleration in that direction equals to zero. Thus, the equation can be simplified to the following: x = v0xt.

An arrow is shot horizontally with an initial speed of 52 m/s. Later it was measured that the point where the arrow hit the target was 30 cm below the launch level. What is the horizontal distance between the target and the launch point?

Hint 1:

Projectile motion problems can be solved by analyzing the motion in the x and y directions separately.

Remember that the motions along x and y axis are completely independent of each other.

Hint 2/Comment:

In order to find the horizontal distance between the target and the launch point, you must know how long the arrow was in the air. Then use the equation x = v0xt + (½)axt2 to find the horizontal distance. Note that the velocity in the x direction is constant, so the acceleration in that direction equals to zero. Thus, the equation can be simplified to the following: x = v0x t.

Comment:

To find the amount of time the arrow spent in the air, use the equation of motion in the y direction. Solve the equation y = v0y t + (1/2)ay t2 for t, and use this value of time in the equation x = v0x t to answer the question of the problem.

An airplane is traveling 30,000 ft above the ground with a constant speed of 800 km/h. At some instant, a package is released from the airplane. What horizontal distance does the package travel before it reaches the ground?

Hint 1:

Projectile motion problems can be solved by analyzing the motion in the x and y directions separately.

Remember that the motions along x and y axis are completely independent of each other.

Hint 2/Comment:

In order to find the horizontal distance the package will travel, you must know how long the package was in the air. Then use the equation x = v0xt + (½)axt2 to find the horizontal distance. Note that the velocity in the x direction is constant, so the acceleration in that direction equals to zero. Thus, the equation can be simplified to the following: x = v0x t.

Comment:

To find the amount of time the package spent in the air, use the equation of motion in the y direction. Solve the equation y = v0y t + (1/2)ay t2 for t, and use this value of time in the equation x = v0x t to answer the question of the problem.

A soccer ball is kicked horizontally off the cliff with the initial speed of 30 m/s. Find the height of the cliff if the ball travels a horizontal distance of 250 m before it reaches the ground.

Hint 1:

Projectile motion problems can be solved by analyzing the motion in the x and y directions separately.

Remember that the motions along x and y axis are completely independent of each other.

Hint 2/Comment:

In order to find the height of the cliff, you must know how long the ball spends in the air.

The height of the cliff can be found using the formula y = v0y t + (½)ay t2. Note that the initial velocity of the ball in the y direction is zero.

Comment:

The time the soccer ball spends in the air can be found using the formula x = v0xt + (1/2)axt2, which simplifies to x = v0x t because ax of the soccer ball equals to zero. Use this value of time in the equation y = v0y t + (1/2)ay t2 to find the height of the cliff.

A monkey sits on a tree 12 m above the ground. It throws a banana to another monkey that is a horizontal distance of 5.7 m away from the base of the tree on the ground. What must be the horizontal initial speed of the banana so that the receiving monkey catches it without having to move?

Hint 1:

Projectile motion problems can be solved by analyzing the motion in the x and y directions separately.

Remember that the motions along x and y axis are completely independent of each other.

Hint 2/Comment:

In order to find the initial horizontal speed, you must know how much time the banana spends in the air.

In order to find the initial horizontal speed of the banana, use the equation x = v0xt + (1/2)axt2, which simplifies to x = v0x t because ax of the banana equals to zero.

Comment:

Use the equation y = v0y t + (½)ay t2 to find how much time the banana spends in the air. Then use this value of time in the equation x = v0x t to solve for v0x (the initial horizontal speed of the banana).

A body thrown horizontally from some height with the speed of 25 m/s lands a horizontal distance of 97 m away from the launch point. What angle does the velocity vector of the body make with the horizontal just before the body hits the ground?

Hint 1/Comment:

Projectile motion problems can be solved by analyzing the motion in the x and y directions separately.

Remember that the motions along x and y axis are completely independent of each other.

Hint 2/Comment:

The velocity vector of the body just before it hits the ground can be decomposed into x- and y- components (see picture). Angle  is the angle the velocity vector makes with the horizontal.

Magnitudes of vectors vx and vy can be found using the equations for motion with constant acceleration. Then  can be calculated as follows: tan  = vy/ vx and  = tan-1(vy/ vx).

Comment:

Use the equation y = (1/2)ayt2 to find the time the body spends in the air. Then plug in this value of time in the equation x = vx t to find vx. After that substitute the same value of time in the equation vy = ayt to find the vertical component of the velocity right before the body reaches the ground.

A small object is thrown horizontally from a high cliff with the initial speed of 34 m/s. How much time will elapse until the speed of the object is exactly 2 (1.5, 2.5, 3, 3.5) times greater than the magnitude of the vertical component of its velocity.

Hint 1/Comment:

By definition, the speed of an object thrown horizontally is equal to:

Hint 2/Comment:

Likewise, we can set up an equation:

, where .

Comment:

Solving for t, we get:

A spherical object is thrown horizontally from point A that is located 5 m above the ground. The object lands on the ground at point B. What angle (in degrees) does the displacement vector make with the horizontal if the initial speed of the object is 7.6 m/s?

Hint 1/Comment:

Projectile motion problems can be solved by analyzing the motion in the x and y directions separately.

Remember that the motions along x and y axis are completely independent of each other.

Hint 2/Comment:

The displacement vector AB can be broken down into x and y components, which can be found using the equations for motion with constant acceleration.

Having found the x component of vector AB, calculate angle  as follows: tan  = y / x, thus  = tan-1(y / x), where x and y are the horizontal and vertical distances traveled by the object respectively.

Comment:

Use the equation y = (1/2)ayt2 to find the time the object spends in the air. Then plug in this value of time in the equation x = vx t to find x.

III. Projectile motion: Understanding (Intuitive) Problems

Consider the trajectory of a projectile shown below. What is the magnitude of the x (y) component of the acceleration at point A (B, C)?

Comment:

Velocity in the x direction is constant throughout an object’s trajectory, thus, the x- component of the acceleration is zero at any point.

Gravity is the only accelerating force acting in the y direction on a projectile in flight. Thus, the acceleration in the y direction is constant and is equal to 9.8 m/s2. Acceleration due to gravity is the same at any point of the trajectory.

The trajectory of a projectile is shown below. Which points on the trajectory are characterized by equal values of vertical speed?

a)A and B only

b)A, B, and C

c)C and D only

d)D and E only

e)A and E only

f)B and D only

g)A, C, and E

h)A and D only

Comment:

The path of a projectile follows the shape of a symmetric parabola. Thus, the absolute values of the slopes, which represent the speeds of the projectile, are equal at the points which are symmetrical to each other on the parabola.

The trajectory of a projective is shown below. Which points on the trajectory are characterized by equal values of acceleration in the x (y) direction?

a)A and B only

b)A, B, and C

c)A, B, C, and D

d)A and C only

e)A and D only

Comment:

The velocity in the x direction is constant and the acceleration is equal to zero. Thus, at every point on the graph the acceleration will be zero.

The acceleration in the y direction is due to gravity. It is also constant at 9.8 m/s2.

A ball thrown at some angle above the horizontal reaches point A, the maximum height, in 5.4 s. How much time will it take the ball to fall from the maximum height back on the ground?

Comment:

Because the path of a projectile follows the shape of a symmetric parabola, the time it takes the projectile to travel the first half of its course is equal to the time it takes the projectile to travel the second half.

An object is launched at some angle above the horizontal. The horizontal distance that the object travels before it hits the ground depends on the…

a)object’s mass

b)material the object is made of

c)object’s volume

d)launch angle

e)horizontal component of the acceleration

Comment:

The equation for the range of a projectile is given by R = v02sin (2Ө)/g. Thus, the only choice given in the question that affects the range of a projectile is the launch angle.

IV. Projectile motion: General Launch Angle

A ball is launched from the ground at an angle of 65o above the horizontal. It reaches the ground 47.0 m away from the launch point. What is the horizontal component of the ball's final speed?

Hint 1:

Projectile motion problems with a general launch angle can be solved by analyzing the motion in the x and y directions separately, just as the projectile problems with a launch angle equal to zero. Because acceleration is constant, the equations for constant acceleration can be used with a slight modification to account for the launch angle of the projectile. The only term that changes in the equations is the velocity: vx=v0 cos andvy = v0 sin . Then the speed is calculated as follows: .

Hint 2:

In order to find the horizontal component of the ball's final velocity, you need to know the initial velocity.

Hint 3:

Since the problem only gives the horizontal range and the launch angle, you can use the equation for the range of a projectile to find the initial velocity:

Comment:

Solve this equation for v0 to find the initial speed of the ball and then use the equation vx=v0 cos to find the horizontal component of v0.

An object is thrown at some angle to the horizontal with the initial speed of 10 m/s. What is the maximum height that the object reaches above the release level if 0.5 s later its speed is 7 m/s?

Hint 1/Comment:

Knowing the definition of speed, we can write the following:

Hint 2/Comment:

Now let’s simplify the system of equations written earlier. Since the horizontal component of the acceleration of the projectile is zero, v0x = v1x. Also, it is true that

, where  is the launch angle and g is the acceleration due to gravity.

Hint 3/Comment:

Substituting those equations in the initial system of equations, we get:

Subtracting one equation of the system from the other and solving for sin (we get:

Hint 4/Comment:

However, we still need to calculate the maximum height that the projectile reaches.

Maximum height can be found using the following formula:

Comment:

Finally, we can obtain the answer by substituting the equation for sin (into the equation for hmax.

A body is thrown at an angle of above the horizontal with the initial speed of . How much time will elapse until the body moves at an angle of above the horizontal?

Hint 1/Comment:

Hint 2/Comment:

The question is asking for the time when the velocity vector is at a specified angle . Remember that the x and y components of a vector form a right triangle. Use trigonometric functions to find a relationship between and the x and y components of the velocity vector.

Hint 3/Comment:

Because the horizontal speed is constant, we can calculate it using the data from the text of the problem: vx=v0 cos .

By definition, the vertical speed of the body equals: vy = v0y – gt, where v0y = v0 sin , thus, vy = v0 sin  – gt.

Comment:

Using the relationships tan  = vy / vx (obtained from the right triangle that the vectors v0, vx, and vy form), vx=v0 cos and vy = v0 sin  – gt, we get:

Solve the above equation for t to answer the question of the problem.

A small rock is thrown with the initial speed of 47 m/s in the direction 70o above the horizontal. What is the maximum height that the rock reaches?

Comment:

Use the equation for the maximum height of a projectile given by ymax = v02sin22gto solve for the maximum height that the rock reaches.

An object is launched from the ground level at the angle of 15o above the horizontal with the initial speed of 30 m/s. What is the horizontal distance that the object travels before it hits the ground?

Comment:

Use the equation for the range of a projectile given by R =v02 sin 2/g to solve for the horizontal distance the object travels before it hits the ground.

A body is thrown from the ground level in the direction of 25o above the horizontal. It hits the ground some time later. Find , the ratio of the range (the horizontal distance that the body travels) to the maximum height that it reaches.

Hint 1:

Use respective equations for R and ymax to write down the ration R / ymax. Simplify this expression by canceling out identical terms in the numerator and the denominator.