Question

A 1.0 kg football is given an initial velocity at ground level of 20.0 m/s [37° above horizontal]. It gets blocked just after reaching a height of 3.0 m.

(a) What is the velocity of the football when it first reaches a height of 3.0 m above the level ground?

(b) What horizontal distance has the ball travelled when it first reaches a height of 3.0 m above ground?

Answer 1

a) velocity can be divide into two components

vx = 20cos37 = 16 m/s

vy = 20 sin37 = 12 m/s

it reaches the height of 3m

v^2 =vy^2 - 2gh

v^2 = 144 - 2*9.8*3

v = 9.23 m/s

Total velocity = sqrt ( 9.23^2 + 16^2 ) = 18.47 m/s

b) time it taken

9.23 = 12 - 9.8*t

t = 0.28 sec

Horizontal distance = s = ut

s = 16*0.28 = 4.48 m