Question

When a ball is thrown up vertically with velocity ${V_0}$ it reaches a maximum height of $h$ . If one wishes to triple the maximum height then the ball should be thrown with velocity.
(A) $\sqrt 3 {V_0}$
(B) $3{V_0}$
(C) $9{V_0}$
(D) $\dfrac{3}{2}{V_0}$

Answer 1

Hint : Here, we know that when the ball is thrown with some velocity vertically upwards at maximum height it becomes zero and comes back down on ground with some velocity. Here, we have to use the formula for maximum height in vertically upward motion. Maximum height means final velocity is zero.
The useful formula is: ${H_{\max }} = \dfrac{{{u^2}}}{{2g}}$
Where, ${H_{\max }}$ is maximum height reached by the ball thrown upward.
$u$ initial velocity, since final velocity is zero we are not using it in the formula.
$g$ is acceleration due to gravity.

Complete Step By Step Answer:
Here, in the above given question the initial velocity of the ball is ${V_0}$ and maximum height is $h$
Thus, by using the formula for maximum height is as follows:
${H_{\max }} = \dfrac{{{u^2}}}{{2g}}$
By placing all the given data in the above formula we have,
$h = \dfrac{{{V_0}^2}}{{2g}}$ …. $(1)$
This is the maximum height of the ball when initial velocity is ${V_0}$
But, when the maximum height is tripled the initial maximum height when the velocity was ${V_0}$ there will be a change in velocity too. Thus, we have to use the same formula to calculate the initial velocity when the height is tripled that of $h$ .
Let initial velocity of the ball be $u'$ , by placing this value we get
$3h = \dfrac{{u{'^2}}}{{2g}}$
$\Rightarrow u{'^2} = 3h \times 2g$
$\Rightarrow u' = \sqrt {3(2hg)}$
$\Rightarrow u' = \sqrt 3 u$ ….. $\left( {\because u = \sqrt {2hg} } \right)$
But, $u = {V_0}$
Therefore,
$\Rightarrow u' = \sqrt 3 {V_0}$
Thus, the initial velocity to reach the maximum height is $\sqrt 3 {V_0}$
The correct answer is option A.

Note :
Here, in this type of question if the object is thrown vertically upward and it reaches maximum height then at that height velocity becomes zero. Thus, in the formula we cannot use the final velocity because it’s already zero. Be careful about the calculation and take care of the values and terms we are using. Give proper reasoning to your steps.