Question

A person stands at the base of a hill that is a straight incline making an angle φ with the horizontal. For a given initial speed, v0 , at what angle θ (to the horizontal) should objects be thrown so that the distance d they land up on the hill is as large as possible?

Answer 1

Answer:

θ = sin⁻¹$\sqrt{2gd}$

Explanation:

From one of the equations of motion, v² = u² + 2as.......... equation 1

Since the object thrown was moving against gravity, then the acceleration, a would change to -g and the initial velocity u would change to V₀ sin θ because the object is travelling at angle of θ to the horizontal. By inputting all these parameter into equation 1, we would arrive at:

v² = (u sin θ)² - 2gd

(u sin θ)² = 2gd

d = (u sin θ)²/2g

sin² θ = 2gd

sin θ = $\sqrt{2gd}$

θ = sin⁻¹ $\sqrt{2gd}$