JJ drops a rock from a window which is 7.80 m high. Kev is 1.80 m tall and is walking toward the front at 3.20 m/s. How far from the front of the school must Kev be when JJ releases the rock?

Question

JJ drops a rock from a window which is 7.80 m high. Kev is 1.80 m tall and is walking toward the front at 3.20 m/s. How far from the front of the school must Kev be when JJ releases the rock?

acceleration (a) =

Vi =

Vf =

∆d =

∆t=

Solve...

Answer 1

I think you left out part of the question. I assume you want to know from how far Kev must start walking so that the rock hits his head as he approaches the building. When the rock hits Kev's head, it has fallen by

d = gt²/2 = (7.80 - 1.80) m = 6.00 m

The time it fell is

t = √(2d/g) = √[2(6.00)/9.8] s = 1.11 s

Kev must walk a distance equal to

vt = (3.20 m/s)(1.11 s) = 3.55 m