Question

# From the top of a building of height $$40 m$$, a boy throws a stone vertically upwards with an initial velocity of $$10 m {s}^{-1}$$ such that it eventually falls to the ground. At what time will the stone reach the ground? $$\left( g=10m{ s }^{ -2 } \right)$$

A
4s
B
3s
C
2s
D
1s

Solution

## The correct option is A $$4 s$$$$s=−40m,u=+10m/s,a=−10m/s^2$$Now $$s=ut+(1/2)at^2$$=> $$−40=10t+1/2×(−10)t^2$$=> $$t^2−2t−8=0$$=> $$(t+2)(t−4)=0$$=> $$t=−2,4s$$The negative value of t is not possibleHence the ball will hit the ground after $$4 s$$Physics

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