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A stone is thrown directly upward with an initial speed of 4 m/s from a height of 20 m. After what time interval does the stone strike the ground

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Given :

( Let , take upward direction +ve and downward direction -ve )

Initial speed of stone , u = 4 m/s .

Height , h = 20 m .

To Find : Time taken to reach ground .

Solution :

We know , by equation of motion .

Displacement is given by :


h=h_o+ut+(gt^2)/(2)\\\\0=20+4t-2t^2\\\\t^2-2t-10=0 ( Here , g = acceleration due to gravity =
-9.8\ m/s^2 .)

Solving above equation , we get :

t = 4.32 s .

Hence , this is the required solution .

User RamenChef
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