A ball is thrown vertically upwords 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
Answers
Answered by
3
h=u²/2g
=49*49/2*9.8
=7*49/2*1.4
=49/2*0.2
=49/0.4
=490/4=245/2=122.5m
t=2u/g=2*49/9.8=2*7/1.4=7/0.2=70/2=35s
=49*49/2*9.8
=7*49/2*1.4
=49/2*0.2
=49/0.4
=490/4=245/2=122.5m
t=2u/g=2*49/9.8=2*7/1.4=7/0.2=70/2=35s
Answered by
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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