Question

A soccer ball is kicked from the top of one building with a height of H1 = 29.2 m to another building with a height of H2 = 10.4 m. (It is not a very smart idea to play soccer on the roof of tall buildings.) The ball is kicked with a speed of v0 = 14.7 m/s at an angle of θ = 73.5° with respect to the horizontal. The mass of a size 5 soccer ball is m = 450 g. What is the speed of the soccer ball, when it lands on the roof of the second bulding? The soccer ball is kicked without a spin. Neglect air resistance.

Answer 1

PROJECTILE

along vertical
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initial velocity v0y = vo*sintheta = 14.7*sin73.5

acceleration ay = -g = -9.8 m/s^2

initial position y0 = H1 = 29.2 m

final position y = H2 = 10.4 m
from equation of motion

vy^2 - voy^2 = 2*ay*(y-yo)

vy^2 - (14.7*sin73.5)^2 = -2*9.8*(10.4-29.2)

vy = -23.8 m/s

along horizontal
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initial velocity v0x = vo*costheta = 14.7*cos73.5

acceleration ax = 0

vx = vox+ a*t

vx = 14.7*cos73.5 = 4.18 m/s

speed v = sqrt(vx^2 + vy^2) = sqrt(4.18^2+(-23.8)^2)

v = sqrt(17.5+566.44)

v=sqrt(583.94)

speed v = 24.16 m/s <<<--------------ANSWER