Question

# A body is dropped from the top of a tower It acquires a velocity $$20 ms^{-1}$$ on reaching the ground. Calculate the height of the tower (Take $$g=10ms^{-2}$$)

Solution

## Given,$$u=0m/s$$$$v=20m/s$$$$g=10m/s^2$$From 3rd equation of motion,$$2gh=v^2-u^2$$$$h=\dfrac{v^2-u^2}{2g}$$$$h=\dfrac{20\times 20-0\times 0}{2\times 10}$$$$h=20m$$The height of the tower is $$20m$$.Physics

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