Question

A girl kicks a football from the ground. 1.52 s later, it hits the side of a building 18.24 m away, at a height of 2.36 m above the ground.

a) Calculate the speed of the football at the instant it left the ground, assuming air resistance is negligible.

b) Calculate the direction of the football's velocity at launch; give answer as an angle measured relative to the horizontal.

c) Calculate the speed of the football at impact.

d) Was the football rising or falling at the instant it hit the building? Explain.

Answer 1

PROJECTILE

1)

along horizontal
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initial velocity v0x = v0*costheta

acceleration ax = 0

initial position = xo = 0

final position = x = 18.24 m

time T = 1.52 s

displacement = x - x0

from equation of motion

x - x0 = v0x*T+ 0.5*ax*T^2

18.24 = vox*1.52

vox = 18.24/1.52 = 12 m/s

along vertical
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initial velocity v0y = v0*sintheta

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

initial position y0 = 0

final position y = 2.36

from equation of motion

y-y0 = v0y*T + 0.5*ay*T^2 .........(2)

2.36 = voy*1.52 - 0.5*9.8*1.52^2

voy = 9 m/s

a)

speed vo = sqrt(vox^2+voy^2) = sqrt(12^2+9^2) = 15 m/s

(b)

direction = tan^-1(Voy/vox) = 36.8

(c)

vy = voy + ay*T

vy = 9 - 9.8*1.52 = -5.896 m/s

vx = vox + ax*T

vx = 12 m/s

speed v = sqrt(vx^2+vy^2) = 13.4 m/s

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(d)

vy is negative

the football is falling at the instant it hit the building