Question #5326

A 100g stone is thrown straight upwards at 19 m/s from the top of a 75 m building.
(a) What is the Stone's velocity as it hits the ground?
(b) What is the maximum height of the stone?
(c)What is the stone's kinetic energy when the gravitational potential energy is 30% less than when it started?
(d) What is the stone's gravitational potential energy 1.0 s before it touches the ground?

Expert's answer

m=100 kg

h=75m

v_{0}=19m/s

a) When the stone hits the ground it`s the gravitational potential energy is equal zero, so let`s use the law of energy conservation:

b) The maximum height of the stone is achieved when the stone`s kinetic energy is equal zero. So let`s use the law of energy conservation:

c) Let`s use the law of energy conservation:

d) We need use the time of the fall

Answer:

a) 42.8 m/s

b) 93.4 m

c) 40.1 kj

d) 39.2 kj

h=75m

v

a) When the stone hits the ground it`s the gravitational potential energy is equal zero, so let`s use the law of energy conservation:

b) The maximum height of the stone is achieved when the stone`s kinetic energy is equal zero. So let`s use the law of energy conservation:

c) Let`s use the law of energy conservation:

d) We need use the time of the fall

Answer:

a) 42.8 m/s

b) 93.4 m

c) 40.1 kj

d) 39.2 kj

## Comments

## Leave a comment