You throw a 20-n rock vertically into the air from ground level. you observe that when it is a height 15.4 m above the ground, it is traveling at a speed of 24.2 m/s upward.

I believe there are two things which we asked to find here:

1. its speed just as it left the ground

2. find its maximum height

Solutions:

1. We use the formula:

ΔKE = - ΔPE

where KE is kinetic energy = ½ mv^2, and PE is potential energy = m g h, Δ = change

Therefore:

½ m (v2^2 – v1^2) = m g (h1 – h2)

at initial point, point 1: h1 = 0, v1 = ?

at final point, point 2: h2 = 15.4 m, v2 = 24.2 m/s

½ (24.2^2 – v1^2) = 9.8 (0 – 15.4)

585.64 – v1^2 = -301.84

-v1^2 = 887.48

v1 = 29.8 m/s

So the rock was travelling at 29.8 m/s as it left the ground.

2. The maximum height (hmax) reached is calculated using the formula:

v1^2 = 2 g hmax

Rewriting in terms of hmax:

hmax = v1^2 / 2 g

hmax = (29.8)^2 / (2 * 9.8)

hmax = 45.3 m

Therefore the rock reached a maximum height of 45.3 meters.

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