A 2 kg body is dropped from a height of 3 m. Calculate:

to. The speed of the body when it is 1 m high

b. The speed of the body when it reaches the ground

A 2 kg body is dropped from a height of 3 m. Calculate:

to. The speed of the body when it is 1 m high

b. The speed of the body when it reaches the ground

Answer:a) 6.26 m/s

b) 7.67 m/s

Explanation:The potential energy at height h0 is initially …

PE0 = mgh0

At height h1, the potential energy is …

PE1 = mgh1

The difference in potential energy is converted to kinetic energy:

PE0 -PE1 = KE1 = (1/2)m(v1)^2

Solving for v1, we have …

mg(h0 -h1) = (1/2)m(v1)^2

2g(h0 -h1) = (v1)^2

v1 = √(2g(h0 -h1))

__

a)When the body is 1 m high, its speed is …v = √(2(9.8)(3 -1)) ≈

6.26 m/s . . . at 1 m high__

b)When the body is 0 m high, its speed is …v = √(2(9.8)(3 -0)) ≈

7.67 m/s . . . when it reaches the ground