The Above Shows the Path of a Ball with Initial Speed 66Ft/Sec

The October, 2008 Problem of the Month

It is well known that a ball thrown at a fixed speed along a flat field will travel farthest if it is thrown at an angle of 45 degrees. (The ball is thrown and caught at the same height, and we neglect air resistance.)

The above shows the path of a ball with initial speed 66ft/sec.

thrown and caught at a height of 5 ft.

Find the angle A at which the ball will travel farthest if it is thrown up a hill that is at a constant angle B with a horizontal. Also find the angle A at which the ball will travel farthest if it is thrown down a hill that is at a constant angle of B with a horizontal.

The above show paths with B = p/8 =22.5 degrees above and below the

horizontal. The initial speed is still 66ft/sec.

Let V be the speed of the throw at angle A. For the uphill throw at time t, the ball is at

x(t) = Vcos(A)t, and y(t) = Vsin(A)t – ½gt2. The ball hits the hill when y/x = tan(B).

This leads to Vsin(A)t – ½gt2 = tan(B)Vcos(A)t, so t = 0 or t = (2V/g)[sin(A) – tan(B)cos(A)].

Plugging the later into x(t) gives x = Vcos(A)(2V/g)[sin(A) – tan(B)cos(A)]. Simplifying with double angle identities we have x = V²/g[sin(2A) – tan(B)cos(2A)-tan(B)]. To find the angle A that maximizes x, we differentiate wrt A,. This leads to tan(B) = -cot(2A), or tan(B) = -tan(π/2 – 2A) = tan(2A – π/2). Therefore B = 2A– π/2 or

A = π/4 + B/2. /

For the downhill throw at time t, the ball is at x(t) = Vcos(A)t, and y(t) = Vsin(A)t – ½gt2. The ball hits the hill when y/x = -tan(B). As above, this leads to tan(B) = cot(2A), or tan(B) = tan(π/2 – 2A) .

Therefore B = π/2 -2A or

A = π/4 - B/2. /