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A player throws a ball upwards with an i...

A player throws a ball upwards with an initial speed of 29.4 m `s^(-1)`.
To what height does the ball rise and after how long does the ball return to the player's hands ? (Take g = 9.8 `ms^(-2)` and neglect air resistance).

Text Solution

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Let the t be time taken by the ball to reach the highest point where height from ground to be S.
Taking vertical upward motion of the ball, we have
`u=-29.4 m//s^(-1), a=9.8m//s^(-2)`, `v=0, S=5, t=2`
As `v^(2)-u^(2)=2as`
`0-(29.4)^(2)=2 times 9.8 times s" (Or)"`
`S=(-(29.4)^(2))/(2 times 9.8)=-44.1m`
Here -ve sign shows that distance is covere in upward direction.
As v = u + at
`therefore 0=-29.4+9.8 times t" (or) "t=(29.4)/(9.8)=3s`
It means time of ascent = 3s Therefore total time after which the ball returns the player.s hand `=3+3=6s`
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