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A projectile is launched from ground level with an initial speed of 57.2 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 :

Answer:

θ = 75°58'

Explanation:

Given,

The initial speed of the of the projectile, u = 57.2 m/s

The initial velocity is above the horizontal plane.

The maximum height the projectile could attain is given by the formula

                                          H = (u²sin²θ)/2g      m

The range of the projectile is given by the formula

                                          R = (u²sin2θ)/g        m

Given that the maximum height of the projectile is equal to its range. Equating both equations H = R

                                (u²sin²θ)/2g = (u²sin2θ)/g

Cancelling out the similar terms

                                      (sin²θ)/2 = sin2θ

                                       sin²θ = 2 sin2θ

                                      sin²θ = 2 x 2 sinθ cosθ

                                      sinθ = 4 cosθ

                                       sinθ/cosθ = 4

                                        tanθ = 4

                                           θ = tan⁻¹ (4)

                                              = 75° 58'

Therefore, the launch angle of the projectile is θ = 75° 58'                                

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