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.
