Problems

Mr. Davis

Regents Physics

1.  A cannonball is fired horizontally from the top of a cliff at 200m/sec. It strikes the ground 4 seconds later.

  1. How high is the cliff?
  2. How far does the cannonball go?
X / y
vi / 200 / 0
vf / 200
a / 9.8
d / 800 / 78.4
t / 4 / 4

d = ½ ay t^2 = 4.9 * 16 = 78.4m

d = vix * t = 200 * 4 = 800m

2.  A projectile is launched at 45 degrees with an initial velocity of 50 m/sec. Find

  1. The maximum height of the projectile
  2. The time of flight of the projectile
  3. The distance the projectile flies (in the x direction).

X / y
vi / 35.4 / 35.4
vf / 35.4
a / 0 / 9.8
d / 255 / 63.8
t / 7.2 / 7.2

viy = 50 * sin45 = 50 * .7071 = 35.4

vix = 50 * cos45 = 50 * .7071 = 35.4

v^2 = 2 a d à viy^2 = 2 a d where

d is the height.

Viy^2/(2a) = d = (35.4)^2/19.6 = 63.8m

d = ½ a t^2 à t = (2d/a)^.5 = (2*63.8/9.8)^.5 = 3.6sec.

This is the time to fall from 63.8 m, so the “hang time” is 7.2sec

The range d = vix * t = 35.4 * 7.2 = 255m.

3.  A ball is thrown downwards at a 30 degree angle and an initial velocity of 3 m/sec. It hits the ground after falling 30m down.

  1. What is its speed in the y direction when it hits the ground?
  2. How long does it take to hit the ground?
  3. How far in the x direction did it travel?
  4. What is its VELOCITY when it hits the ground? (Vector quantity)

x / y
vi / 2.6 / 1.5
vf / 2.6 / 24.3
a / 0 / 9.8
d / 6 / 30
t / 2.33 / 2.33

vix = 3 * cos 30 = 3*0.866 = 2.6

viy = 3 * sin 30 = 3* 0.5 = 1.5

vf2 = vi2 + 2ad

vfy2 = viy2+2ad = 2.25 + 2 * 9.8* 30

= 24.3

dy = viy t+ 1/2 a t^2 = 1.5t + ½ * 9.8 t^2

30 = 1.5t + 4.9t^2 à 0 = 4.9t^2 + 1.5t -30

t = (-1.5 +/- ((1.52 – 4 * 4.9 * (-30)))^.5)/2*4.9 or vfy = viy + at

= 22.8/(2*4.9) (vfy – viy)/a = t = (24.3-1.5)/9.8 = 2.33sec

= 2.33 sec

dx = vxt = (2.6) * 2.33 = 6.05m

Velocity = (vfx2 + vfy2)^.5 = (24.3^2 = 2.6^2)^.5 = 24.44m/sec

Angle is arctan (vfy / vfx) = arctan (9.35) = 83.9 degrees (angle of inclination)

4.  A cannon is fired over the edge of a cliff that is 100m high. The cannonball leaves the cannon with a muzzle velocity of 100m/sec.

  1. How far does the cannonball travel?
  2. How long does it take to hit the ground?

x / y
vi / 100 / 0
vf / 100
a / 0 / 9.8
d / 452 / 100
t / 4.52 / 4.52

dy = viyt + ½ ay t^2

100 = 0 + ½ * 9.8 t^2

t^2 = 2*100/(9.8) = 20.4

t = 4.52

dx = vix * t = 100 * 4.52 = 452m

5.  An astronaut on the moon hits a golf ball with a speed of 90m/sec and at an angle of 37 degrees from the horizontal. Note that the acceleration due to gravity on the moon is approximately 1.6m/sec2.

  1. How far does the ball go?
  2. How high does it get?
  3. How long does it take to hit the ground?

x / y
vi / 71.9 / 54.1
vf / 71.9
a / 0 / 1.6
d / 4862 / 915
t / 67.6 / 67.6

vi = 90 m/sec

vix = 90 cos(37) = 71.9m/sec

viy = 90 sin(37) = 54.1m/sec

a = 1.6 m/sec^2

vfy = viy –at, treating vi as positive and a as negative. At the top of the parabola, vy = 0. Using this as vfy,

vfy = viy –ayt à 0 = viy – at à viy = ayt à viy /ay = t = 54.1/1.6 = 33.8 sec at top of parabola, so full “hang time” is 67.6 sec.

dy = viyt + 1/2ayt^2. Treating ay as negative,

dy = viyt – ½ ayt^2 = 54.1 * 33.8 – ½ * 1.6 * (33.8)^2 = 1829 – 914 = 915m

The distance the ball travels is dx = vix t = 71.9 * 67.6 = 4862m.

6.  A rocket is fired at an angle upwards and hits the ground 60 seconds later after traveling 100km.

  1. With what speed and at what angle was it fired?

x / y
vi / 1667 / 294
vf / 1667
a / 0 / 9.8m/s^2
d / 100km / 4410m (height)
t / 60 / 60

Vix = vi cos A

Viy = vi sin A

dx = vxt à dx/t = vix t = 1667m/sec

Height of the parabola is at 30 seconds.

So dy = vy + ½ a t^2 = ½ a t^2 since vy = 0 at the top of the arc.

dy = 4.9(30)^2 = 4.9*900 = 4410m.

Then viy = at = 9.8 * 30 = 294 m/sec, and the firing angle

A = arctan(294/1667) = 10 degrees.

Vi = (viy^2 + vix ^2)^.5 = 1692 m/sec.

7.  A player kicks a football at an angle of 37o with the horizontal and with an initial speed of 48 m/sec. A second player standing at a distance of 100 m from the first in the direction of the kick starts running to meet the ball at the instant it is kicked. How fast must he run in order to catch the ball before it hits the ground?

x / y
vi / 38.3 / 28.9
vf / 38.3
a
d
t

Vi = 48m/sec

Vix = vi cos 37 = 38.3 m/s

Viy = vi sin 37 = 28.9 m/sec

The height of the ball is viy^2 = 2 a d,

So dy= viy^2/2a = 28.9^2/(2*9.8) = 42.6m

dy = ½ ay t^2 à(2dy/a)^.5 = t = 2.94 at the top of the arc, implying that the hang time is

2* 2.94 = 5.9sec

dx = vix * t = 38.3 * 5.9 = 225m. So the player must run (225 – 100) m in 5.9 sec, or at a speed of 21.1 m/sec

.

8.  A projectile shot at an angle of 60o above the horizontal strikes a building 80 ft away at a point 48 ft above the point of projection.

a.  Find the initial velocity

b.  Find the magnitude & direction of the velocity when it strikes the building

9.  A basketball player releases the ball 7 ft above the floor when he is 30 ft from the basket. The ball goes through the rim of the basket (which is 10 ft above the floor) 1.5 seconds after release. Find

a.  The initial velocity,

b.  The maximum height above the floor reached by the ball.