Answer:
The angle of launch from the horizontal direction is 20.99° .
Explanation:
Let u and θ be the initial speed and angle of projection from the horizontal axis of the object respectively.
The equations for projectile motion are :
H = ( u² sin²θ)/ 2g ......(1)
Here H is maximum height of the projectile motion and g is acceleration due to gravity.
R = ( u² sin2θ)/g .......(2)
Here R is the maximum horizontal displacement of the object.
Rearrange equation (1) in terms of u².
u² = (2gH)/sin²θ
Substitute this equation in equation (2).
R = (2gH sin2θ) / (sin²θ x g)
R = (2H sin2θ)/sin²θ
Using trigonometry property, sin2θ = 2 cosθ sinθ
So, above equation becomes,
R = (2H x 2 cosθ sinθ)/sin²θ
R = (4H cosθ)/sinθ
tanθ = R/4H
θ = tan⁻¹(R/4H)
Substitute 111 m for R and 72.3 m for H in the above equation.
θ = tan⁻¹( 111/ 4 x 72.3 )
θ = tan⁻¹(0.38)
θ = 20.99°
