Question

A projectile from the ground has its direction of motion making an angle π/4 with horizontal at a height 40m. Its intital velocity of projection is 50m/s, the angle of projection is??

a) 1/2cos^-1(-8/25)
b) 1/2cos^-1(8/25)
c) 1/2cos^-1(-4/5)
d) 1/2cos^-1(-1/4)

Answer 1

Answer:

(1/2)Cos⁻¹(-8/25)

Explanation:

Let say Angle of projection = α

velocity of projection is 50m/s,

Vertical speed at time of projection = 50Sinα

Horizontal speed at time of projection = 50Cosα

π/4  = 45°

Let say Velocity is V at 40 m height

50Cosα = VCos45°   => V = 50√2Cosα

VSin45° is vertical velocity = 50√2Cosα ( 1/√2) = 50Cosα

Using formula V² - U² = 2aS

V = VSin45° = 50Cosα  U = 50Sinα  a = -g= -10  S = 40

(50Cosα)² - (50Sinα)² = 2(-10)(40)

=> 50² (Cos²α - Sin²α) = -800

=>  Cos²α - Sin²α = -8/25

=> Cos2α = -8/25

=> 2α  = Cos⁻¹(-8/25)

=> α = (1/2)Cos⁻¹(-8/25)

the angle of projection is (1/2)Cos⁻¹(-8/25)

Answer 2

Explanation:

answer is in attachment... hope it helps