Question

A ball is thrown at 60 degrees and lands in 18.5 seconds what is the velocity of the ball at the start

Answer 1

the initial velocity of the ball is 104.67 m/s.

Step-by-step explanation:

Given;

angle of projection, θ = 60⁰

time of flight, T = 18.5 s

let the initial velocity of the ball, = u

The time of flight is given as;

`T = (2u* sin(\theta))/(g) \\\\2u* sin(\theta) = Tg\\\\u = (Tg)/(2* sin(\theta)) \\\\u = (18.5 * 9.8)/(2* sin(60^0)) \\\\u = 104.67 \ m/s`

Therefore, the initial velocity of the ball is 104.67 m/s.