Question

A fruit bat falls from the roof of a cave. We know that her potential energy was

600 J, and that the roof of the cave was 12.8 meters high. What is her velocity
when she hits the ground?​

Answer 1

v = 15.65 m/s

Step-by-step explanation:

We use conservation of mechanical energy between initial (i) and final (f) states:

Pi + KEi = Pf + KEf

At the top of the cave at the instant the bat starts to fall, there is only potential energy since the bat's velocity is zero.

Pi = m g h = 600 J

and the KEi = 0 J (no velocity)

Knowing the height of the cave's roof (12.8 m) , we can find the mass of the bat:

m = 600 J / (g 12.5) = 4.9 kg

Using conservation of mechanical energy, the final state is:

Pf + KEf = 600 J

with Pf = 0 (just touching the ground)

KEf= 1/2 4.9 (v^2)

and we solve for the velocity:

600 J = 0 + 1/2 4.9 (v^2)

v^2 = 600 * 2 / 4.9 = 244.9

v = 15.65 m/s