Question

A body is projected vertically up with a certain velocity so that it reaches a maximum height of 5m. If the body is projected with double the initial velocity then find the maximum height reached?​

Answer 1

Answer: s=20 m when initial velocity,u is doubled

Explanation:

We know that

v=0 (as the object stops at the highest point )

a=-9.8 m/s^2 (as the object is moving in the opposite direction as that of gravitational force of earth)

s=5 m (given)

by the third equation of motion

$v^2=u^2+2as$

$0^2=u^2+2 * (-9.8)*5$

$-u^2=-98$

$u^2=98$

$u=\sqrt{98}$

$u=7\sqrt{2}$

A.T.Q,

we have to find the distance when u is doubled

So,

$2u=2*7\sqrt{2}\\ = 14\sqrt{2}$

Now we will find the distance by the help of third equation of motion

So,

$v^2=u^2+2as$

$0^2 = (14\sqrt{2})^2+ 2*(-9.8)*s$ (substituted the values from above)

$0=196*2-19.6s$

$0=392-19.6s$

$-392=-19.6s$

$392/19.6=s$(minus and minus on both the sides got cancelled by each other and 19.6 is shifted to the left hand side)

so,

$s=20 m$

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