A boy stands at 78.4 m from a building and throws a ball which just enters a window 39.2 m above the ground. Calculate the velocity of projection of the ball. 

A boy standing at A throws a ball with velocity u at an angle θ with the horizontal, which just enters window W.

As the boy is 78.4 m from the building and the ball enters the window 39.2 m above the ground, therefore:

Maximum height,

u² sin² θ / 2g = 39.2 m … (i)

and horizontal range,

u² sin 2θ / g = 2 × 78.4 m … (ii)

Dividing (i) by (ii), we get:

(u² sin² θ / 2g) × (g / u² sin 2θ) = 39.2 / (2 × 78.4)

⇒ sin² θ / (2 × 2 sin θ cos θ) = 1 / 4

⇒ tan θ = 1 ⇒ θ = 45°

Substituting in (ii), we get:

u² sin 90° / 9.8 = 2 × 78.4

⇒ u = √(2 × 78.4 × 9.8) = 39.2 m/s

Hence, the required velocity of projection is 39.2 m/s at an angle of 45°.