Question

A body is throw up with a velocity 'u'. It reaches maximum height 'h'. If its velocity
of projection is doubled the maximum height it reaches is
​

Answer 1

Answer:

Initial velocity = u

Maximum height = h

By using the formula,

$h = \frac{ {u}^{2} }{2g}$

New u = 2u

Then,

$h = \frac{ {(2u)}^{2} }{2g} \\ = \frac{ {4u}^{2} }{2g} \\ = \frac{ {2u}^{2} }{g}$

Hence ans is 2u^2/g

Thank you.

Hope it hepls.

Please hit a THANKS