A ball thrown up vetically returns to the thrower after 6s. Find the velocity with with it was thrown up, the maximum height it reaches and its position after 4s.
Answers
_/\_Hello mate__here is your answer--
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The ball takes a total of 6 s for its upward and downward journey. (Time of ascent is equal to the time of descent.)
Hence, it has taken 3 s to reach at the maximum height.
v = 0 m/s
g = −9.8 ms−2
Using equation of motion,
v = u + at
⇒0 = u + (−9.8 × 3)
⇒ u = 9.8 × 3 = 29.4 m/s
Hence, the ball was thrown upwards with a velocity of 29.4 m/s.
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Let the maximum height attained by the ball be h.
u = 29.4 m/s
v = 0 m/s
g = −9.8 ms−2 (upward direction)
Using the equation of motion,
s = ut +1/2 gt^2
⇒ hℎ = 29.4 × 3 −1/2 × 9.8 × 32
⇒ ℎh = 44.1 m
Hence, the maximum height is 44.1 m.
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Ball attains the maximum height after 3 s. After attaining this height, it will start falling downwards.
In this case,
u = 0 m/s
Position of the ball after 4 s − 3 s = 1 s can calculate by Using the equation of motion,
s= ut +1/2gt^2
⇒ h= 0 × 1 +1/2 × 9.8 × 12
⇒ h= 4.9 m
Now, total height = 44.1 m This means, the ball is 39.2 m (44.1 m − 4.9 m) above the ground after 4 seconds.
I hope, this will help you.☺
Thank you______❤
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