Question

A ball is thrown vertically upwards with a velocity of 49 m/s
Calculate the maximum height to which it rises and the total time it takes to return to the surface of earth

Answer 1

$formula \: for \: max \: height = \frac{ {u}^{2} }{2g} \\ so \: max \: height = \frac{ {49}^{2} }{2 \times 9.8} \\ = 122.5m$
$total \: time = \frac{2u}{g} \\ = 98 \div 9.8 = 10sec$

Answer 2

_/\_Hello mate__here is your answer--

_________________

v = 0 m/s and

u = 49 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 = 2gh

⇒ 0^2 − 49^2 = 2(−9.8)ℎ

⇒ ℎ =49×49/ 2×9.8 = 122.5

_______________

Let t be the time taken by the ball to reach the height 122.5 m, then according to the equation of motion

v= u+ gt

=>0 = 49 + (−9.8) t

⇒ 9.8 t= 49

⇒ t = 49/9.8 = 5 s

But, Time of ascent = Time of descent

Therefore, total time taken by the ball to return

= 5 + 5 = 10

I hope, this will help you.☺

Thank you______❤

___________________❤