Question

On top of a cliff of height h, a spring is compressed 5m and launches a projectile perfectly horizontally with a speed of 75 m s . It hits the ground with speed 90 m s . How high above the ground was the cliff? (Hint: use energy conservation to make the problem easier!)

Answer 1

Answer:

The height of the cliff is 121.276 m

Explanation:

Given;

initial velocity of the projectile, v₁ = 75 m/s

final velocity of the projectile, v₂ = 90 m/s

spring compression = 5 m

Apply the law of conservation of energy;

mgh₀ + ¹/₂mv₁² = mgh₂ + ¹/₂mv₂²

gh₀ + ¹/₂v₁² = gh₂ + ¹/₂v²

gh₁  - gh₂ = ¹/₂v₂² - ¹/₂v₁²

g(h₀  - h₂) = ¹/₂ (v₂² - v₁²)

h₀  - h₂ = ¹/₂g (v₂² - v₁²)

h₀ = h(cliff) + 5m

when the projectile hits the ground, Final height, h₂ = 0

$h_o - 0 = \frac{1}{2g}(v_2^2-v_1^2)\\\\h_{cliff} + 5= \frac{1}{2g}(v_2^2-v_1^2)\\\\h_{cliff} = \frac{1}{2g}(v_2^2-v_1^2) - 5\\\\h_{cliff} = \frac{1}{2*9.8}(90^2-75^2) - 5\\\\h_{cliff} = 121.276 \ m$

Therefore, the height of the cliff is 121.276 m