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A ball is thrown upwards from the top of a tower `40 m` high with a velocity of `10 m//s.` Find the time when it strikes the ground. Take `g=10 m//s^2`.

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In the problem `u= + 10 m//s, a =-10 m//s^(2) and s =- 40 m ` ( at the point where ball strikes the ground)
Substituting in` s = ut +(1)/(2) at^(2)`
`-40= 10t - 5t^(2) or 5t^(2) -10t -40=0 or t^(2) - 2t -8 =0`
Solving this we have t= 4 s and -2 s. Taking the positive value t = 4s.
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