Time taken by a body thrown up, to reach the maximum height is given by ........................

\frac{u^{2}}{g}

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\frac{2 u}{g}

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\frac{\sqrt{u}}{g}

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\frac{u}{g}

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Solution

The correct option is B $\frac{u}{g}$
Let maximum height be h

At height h, v=0
Now we know that v=u+at
Hence 0=u+(-g)h , as the direction of motion is upwards.
Hence, h= (u/g)

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Time taken by a body thrown up, to reach the maximum height is given by ........................

The correct option is B ugLet maximum height be hAt height h, v=0Now we know that v=u+atHence 0=u+(-g)h , as the direction of motion is upwards.Hence, h= (u/g)

The correct option is B $\frac{u}{g}$
Let maximum height be h

At height h, v=0
Now we know that v=u+at
Hence 0=u+(-g)h , as the direction of motion is upwards.
Hence, h= (u/g)