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
[tex]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[/tex]
Therefore, the height of the cliff is 121.276 m
