A ball is thrown horizontally from the top of
a building 120 m high. The ball strikes the
ground 63 m horizontally from the point of
release. What is the speed of the ball just before it
strikes the ground?
Answer in units of m/s.
1
Expert's answer
2019-10-07T10:47:18-0400
Let the height of the building be H=120m and the horizontal travel distance L=63m.
The equations of motion are as follows:
vx(t)=v0x, vy(t)=v0y−gt=−gt (no initial vertical component of velocity),
H=2gt2, L=v0xt .
From the last two equations, t=g2H, v0x=tL=2HLg. The speed before hitting the ground is v=vx2+vy2=v0x2+(gt)2=2HgL2+2gH≈50.2sm.
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