Answer:
a) v₀ = 39.83 v / s , b) y = 111.94 m
Explanation:
a) Let's use the kinematic relations for this problem
v² = v₀² - 2 g y
The speed at the base of the cliff is 30 m / s, the height y = 31 m
v₀² = v² + 2 g y
v₀ =√ (30² + 2 9.8 31)
v₀ = 39.83 v / s
b) at the point of maximum height the speed is zero
0 = v₀² - 2 g (y –y0)
y = y₀ + v₀² / 2g
.y = 31 + 39.83²/2 9.8
y = 111.94 m
