Question

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Answer 1

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The ball is thrown up and it returns in 6sec,

Time of accent = Time of decent

So, time to reach the highest point =6/2=3s

By 2nd equation of motion

$s = ut + \frac{1}{2} at {}^{2}$

To calculate height, consider motion of the ball from highest point to the ground

$h = \frac{1}{2} at {}^{2} = \frac{1}{2} \times 10 \times 3 {}^{2} = 45m$

To calculate projection velocity, consider motion of the ball from projection to return to the point of projection.

$0 = ut + \frac{1}{2} at {}^{2}$

$u = 10 \times \frac{6}{2} = 30ms {}^{ - 1}$

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