Question

a ball is thrown vertically upward calculate
a: the total time to which it rises ( velocity upwards 49m/s)
b: the total time to which it take to return to earth

Answer 1

the initial upward velocity is given as 49 m/s
the final velocity at maximum height will be zero.
v = u + at
here 
v=0
u = 49 m/s
a = -9.8 m/s^2

0= 49 - 9.8t
t = 49/9.8 = 5 seconds.
it takes 5 seconds to rise.

and similarly while falling, it takes 5 seconds.
so total time is 10seconds

Answer 2

_/\_Hello mate__here is your answer--

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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 = 2h

⇒ 0^2 − 49^2 = 2(−9.8)ℎ

⇒ ℎ =49×49/ 2×9.8 = 122.5

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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

⇒t 9.8 = 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______❤

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