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A ball throuwn up is cautght by the thro...

A ball throuwn up is cautght by the thrower after 4s. How did it go and with what velocity was it thrown ? How far was it below the highest point 3s after it was thrown?

Text Solution

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As time of ascent=time of descent
`:.` Time taken by the ball to reach the highest point=2s
For upward motion of the ball :
u=?, v=0, t=2s, g=-9.8`ms^(-2)`
As v=u+"gt"
`:." "0=u-9.8xx2` or u`=9.6ms^(-1)`
Maximum height attained by the ball is given by
`s=ut+1/2"gt"^2=19.6xx2+1/2xx(9.8)xx2^2=19.6m`
Displacement of the ball in 3s
`s=19.6xx3+1/2xx(-9.8)xx3^2=58.8-44.1=14.7m`
Distance of the ball from the highest point 3 s after it was thrown
`=19.6-14.7=4.9m`
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