Question

You throw a 20-N rock vertically into the air from ground level. You observe that when it is a height 15.9 m above the ground, it is traveling at a speed of 24.8 m/s upward.

A.Use the work-energy theorem to find its speed just as it left the ground.

B.Use the work-energy theorem to find its maximum height.

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Weight of the rock = W = 20 N

Mass of the rock = m

Speed of the rock just as it left the ground = V₀

Speed of the rock at a height 15.9 m above the ground = V₁ = 24.8 m/s

Height of the rock where the speed of the rock is 24.8 m/s = H₁ = 15.9 m

The potential energy and the kinetic energy of the rock on the ground is equal to the potential energy and kinetic energy of the rock at any height.

mV₀²/2 = mgH₁ + mV₁²/2

V₀² = 2gH₁ + V₁²

V 2gH1 V2

Vo (2)(9.81) (15.9)(24.8)2

V₀ = 30.446 m/s

Maximum height reached by the rock = H₂

By conservation of energy the potential energy of the rock at maximum height is equal to the potential energy and kinetic energy of the rock when it is at 15.9 m above the ground.

mgH₂ = mgH₁ + mV₁²/2

На — Ні + 2g

(24.8)2 H2= 15.92)(9.81) (2) (9.81)

H₂ = 47.247 m

A) Speed of the rock just as it left the ground = 30.446 m/s

B) Maximum height reached by the rock = 47.247 m