Question

A projectile is launched from ground level with an initial speed of 48.7 m/s. Find the launch angle (the angle the initial velocity vector is above the horizontal) such that the maximum height of the projectile is equal to its horizontal range. (Ignore any effects due to air resistance.)

Answer 1

Answer:

∅ = 75.96º

Explanation:

vi = 48.7 m/s

If Xmax = Ymax

Knowing that

Xmax = vi*Cos∅*t

t = 2*vi*Sin ∅ / g     ⇒   Xmax = vi*Cos∅*(2*vi*Sin ∅ / g) = vi²*Sin (2∅) / g  (I)

and

Ymax = vi²*(Sin ∅)² / (2*g)    (II)

then we get (I) = (II)

vi²*Sin (2∅) / g = vi²*(Sin ∅)² / (2*g)  ⇒  

2*Sin (2∅) = (Sin ∅)²  ⇒   2*(2*Sin ∅* Cos ∅) = (Sin ∅)²  ⇒  4*Cos ∅ = Sin ∅

⇒  Sin ∅ / Cos ∅ = tan ∅ = 4   ⇒  ∅ = Arctan (4) = 75.96º