Question

A ball is thrown vertically upwards with a velocity of 49m/s. Calculate (i) the maximum height to which it rises, (ii) the total time it takes to return to the surface of the earth,its position after 6 seconds

Answer 1

u = 49 m/s
v = 0
a = -9.8 m^2

(i)
t = v-u/a
t = -49/-9/8
t= 5 secs

v^2-u^2 = 2as
0-2401 =  2 x -9.8 x s 
2401 = 19.6 x s
s = 2401/19.6
s = 122.5 metres

(ii) as Time of reaching the highest point = Time of reaching the surface of earth 
Therefore, total time = 5 + 5 = 10 secs

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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