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
Therefore, the height of the cliff is 121.276 m
Answer:
The angle above the horizontal at which the pitcher throws the ball determines the distance the ball travels before returning to the height at which it was thrown
Explanation:
The baseball is thrown as a projectile and the range, 'R', of the baseball which is the distance the baseball travels before the height above the ground returns to the initial height is given given as follows;
Where;
R = The range of the baseball = The horizontal distance away from the pitcher the ball reaches
u = The initial velocity with which the baseball was thrown
θ = The angle above horizontal a baseball pitcher throws the ball
g = The acceleration due to gravity ≈ 9.81 m/s²
From the the equation, when θ = 0, sin(θ) = sin(0) = 0 and the ball does not cover any horizontal distance before going lower than the height at which it was thrown, therefore, for the ball to travel further, the angle of launch, θ has to be larger than 0.