A ball is thrown vertically upward with a velocity of 20 ms-1 from a top of a multistory building. The height of the point where the ball is thrown 25 m from the ground. How long will it be before the ball hits the ground? Take g=10ms-2

Answers (1)

To calculate the time to reach maximum height, we can use v=u+at,

\begin{array}{l} 0=20+(-10) t \\ t=2 s \\ s=u t+0.5 a t^{2}=20(2)+0.5(-10)(2)^{2}=20 m \end{array}

Thus, the total distance for maximum height is 45 m

\begin{array}{l} s=u t+0.5 a t^{2} \\ 45=0+0.5(10)\left(t^{\prime}\right)^{2} \\ t^{\prime}=3 s \\ \text {Total time }=3+2=5 \mathrm{s} \end{array}

 

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