Question

A football thrown over a level field travels a horizontal distance of 17 m before hitting the
ground. The point of release is 1.5 m above the ground and the projection angle is 16 degrees. What is the ball’s initial speed?

Answer 1

Given info : A football thrown over a level field travels a horizontal distance of 17 m before hitting the ground. The point of release is 1.5 m above the ground and the projection angle is 16°C.

To find : initial speed of football.

Solution : let u is initial speed of ball.

horizontal speed of ball = ucos16°

vertical speed of ball = usin16°

Horizontal distance = 17 m

using formula, x = u_x × t

⇒17 = ucos16° × t

⇒t = 17/ucos16° ......(1)

Vertical distance = 1.5 m

Using formula, y = u_yt - 1/2 gt²

⇒1.5 = usin16° × 17/ucos16° - 1/2 × 10 × (17/ucos16°)² [ from eq (1).]

⇒1.5 = 17tan16° - 5 × 289/u²cos²16°

⇒1.5 = 17 × 0.28 - 5 × 289/u²(0.96)

⇒-3.26 = - 1445/u²(0.96)

⇒u² = 1445/(3.26 × 0.96) = 461.7

⇒u = 21.5 m/s

Therefore the initial speed of ball is 21.5 m/s