Answer:
θ = sin⁻¹[tex]\sqrt{2gd}[/tex]
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 = [tex]\sqrt{2gd}[/tex]
θ = sin⁻¹ [tex]\sqrt{2gd}[/tex]