Answered

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

Sagot :

onyebuchinnaji

Answer:

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

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;

[tex]T = \frac{2u\times sin(\theta)}{g} \\\\2u\times sin(\theta) = Tg\\\\u = \frac{Tg}{2\times sin(\theta)} \\\\u = \frac{18.5 \times 9.8}{2\times sin(60^0)} \\\\u = 104.67 \ m/s[/tex]

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

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