Question

For which angle of projection the range and its maximum height will be equal?
(A) θâ = 45°
(B) θâ = tanâ»Â¹ (4)
(C) θâ = tanâ»Â¹ (1/4)
(D) θâ = 30°

Answer 1

Answer:

Tan⁻¹(4)

Explanation:

Let say Angle of Projection = α

Vertical Velocity = VSinα

Time to to reach max height = VSinα/g   ( using V = U + at)

a = g

Max height = V²Sin²α/2g   ( using V² - U² = 2aS)

Horizontal Velocity = VCosα

Time of Flight = 2 * VSinα/g

Horizontal Range =  VCosα *2 * VSinα/g

V²Sin²α/2g  = VCosα *2 * VSinα/g

=> Tanα = 4

=> α = Tan⁻¹(4)