Question

A ball is thrown upward with a velocity of 4.9 m/s calculate the maximum hight to reach the risers and the total time takes to return to the surface of earth.

Answer 1

initial velocity of ball, u = 4.9 m/s

let the maximum height = h

total time it takes to return or time of flight = t

$h = \frac{ {u}^{2} }{2g} = \frac{ {4.9}^{2} }{2 \times 9.8} = 1.225 \: m$

$t = \frac{2u}{g} = \frac{2 \times 4.9}{9.8} = \frac{9.8}{9.8} = 1 \: sec$


maximum height = 1.225 m

time of flight = 1 second

Answer 2

_/\_Hello mate__here is your answer--

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v = 0 m/s and

u = 4.9 m/s

During upward motion, g = − 9.8 m s^−2

Let h be the maximum height attained by the ball.

using

v^2 − u^2 = 2h

⇒ 0^2 − 4.9^2 = 2(−9.8)ℎ

⇒ ℎ =4.9×4.9/ 2×9.8 = 1.225

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Let t be the time taken by the ball to reach the height 1.225 m, then according to the equation of motion

v= u + gt

=>0 = 4.9 + (−9.8) t

⇒t 9.8 = 4.9

⇒ t= 4.9/9.8 =0.5 s

But, Time of ascent = Time of descent

Therefore, total time taken by the ball to return

= 0.5 + 0.5 = 1.0

I hope, this will help you.☺

Thank you______❤

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