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A ball thrown up vertically returns to t...

A ball thrown up vertically returns to the thrower after 6s. Find
(a) the velocity with which it was thrown up.
(b) the maximum height it reaches and
(c) its position after 4s.

Text Solution

Verified by Experts

The correct Answer is:
(a) `29.4ms^(-1)` (b) `44.1m` (c ) 39.2m
Total time taken =6s
time of ascent =time of descent `=6/2=3sec`
(a) For the upward motion of the ball, v=0,
t=3 sec, `g=-9.8 ms^(-2)`
`v=u+gt`
`0=u-9.8 xx 3 therefore u=29.4ms^(-1)`
(b) `s=ut+1/2 gt^(2)`
`=29.4 xx 3 +1/2 xx(-9.8) xx (3)^(2)`
`=88.2-44.1`
s=44.1m
(c) Position of ball aftet 4 sec.
`s=ut+1/2 gt^(2)`
`=29.4 xx 4+1/2 xx (-9.8) xx (4)^(2)`
`=117.6-78.4=39.2m`
The ball is at a height of 39.2 m from the ground.
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