Question

Mark is standing of the edge of the roof of a 25.0 m tall building. He kicks a soccer ball at a velocity of 14.0 m/s at an angle of 37.0° above the horizontal.

a) How long will the ball be in the air (assume no air resistance and no buildings in the way to interfere with the path of the ball)?

b) What will be the maximum height of the ball?

c) With what velocity will the ball hit the ground?

Answer 1

using the equation we have

Y = Uyt - 0.5gt^2

-25 = 14 sin 37*t - 0.5*9.8*t^2

4.9t^2 - 8.42t - 25 = 0

t = 8.42 + sqrt(8.42^2 + 4*4.9*25) / 2*4.9

= 3.27 s

so the ball remains in the air for 3.27s

b) Vy^2 - Uy^2 = 2gH

at maximum height Vy = 0

0 - (14 sin 37)^2 = 2*-9.8*H

H = 3.62 m

so maximum height above the ground = 25 + 3.62 = 28.62 m

c) since there is no acceleration in x-direction

Vx = Ux = 14 cos 37 = 11.18 m/s

Vy = Uy - gt

= 14 sin 37 - 9.8*3.62 = -27 m/s

velocity with which it hits the ground = sqrt(Vx^2 + Vy^2) = sqrt(11.18^2 + (-27)^2 ) = 29.27 m/s