Question

1.) A golf ball is launched at an angle of 60 o to the level ground and at a speed of 39.2 m/s. Find (a) The time it takes to reach its maximum height. (b) The maximum height it attains. (c) The time it takes to fall back to the ground. (d) The horizontal range.

Answer 1

a] t = u sin theta/g

= 39.2 sin 60 degree/9.8

= 3.46 s

b] max height, h = [u sin theta]^2/2g

= [ 39.2*sin 60 degree]^2/[2*9.8]

= 58.8 m

c] Same time as it takes to reach its maximum height

t = 3.46 s

d] Horizontal Range, R = u^2 sin 2theta/g

= [39.2]^2*sin 120 degree/9.8

= 135.79 m answer