A stone dropped from the top of a 80m high building strikes the ground at 40 m/s

A

Explanation:

Given that a stone dropped from the top of a 80m high building strikes the ground at 40 m/s after falling for 4 seconds. The stone's potential energy with respect to the ground is equal to its kinetic energy. (use 9 - 10 m/s)

O at the moment of impact

2 seconds after the stone is released after the stone has fallen 40 m

when the stone is moving at 20 m/s

At the top of the hill, the P.E = mgh

P.E = 10 × 80 × m

P.E = 800m

At the moment of impact, K.E = 1/2mv^2

K.E = 1/2 × 40^2 × m

K.E = 1/2 × 1600 × m

K.E = 800m

Since both P.E and K.E are the same, we can therefore conclude that the stone's potential energy with respect to the ground is equal to its kinetic energy at the moment of impact.

The correct answer is option A.

before touching it is lower but after touching it's slightly higher because body warmth is transferred to the material through conduction