Question

A ball is thrown up vertically returns to the thrower after 6sec. Find the velocity with which it was thrown up the maximum height it reaches and its position after 4sec.

Answer 1

Here take a look at your answer... hope it helps..

Answer 2

_/\_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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