Question

Two stones are thrown vertically upwards simultaneously with initial velocities ${u_1}$ and ${u_2}$ respectively. What is the ratio of maximum heights reached by them?

Answer 1

Hint: To find the maximum heights reached by the two stones, we need to apply one of the equations of motion. Then using that equation of motion we can find out the expressions for the distances travelled by two stones.
Formula used: In this solution we will be using the following formula,
$\Rightarrow {v^2} = {u^2} + 2as$
$\Rightarrow {v^2} = {u_1}^2 + 2a{h_1}$
$\Rightarrow {v^2} = {u_2}^2 + 2a{h_2}$
Where, the final velocities of the two stones given by $v$ are zero, as the two stones reach maximum height. The initial velocities of the stones are ${u_1}$ and ${u_2}$. The distances covered by the stones are ${h_1}$ and ${h_2}$ respectively.

Complete step by step solution:
Let us consider, the final velocity of the two stones be $v$, which is zero, as the two stones reach maximum height. The initial velocities of the stones are ${u_1}$ and ${u_2}$.
We know the equation of motion which involves distance, initial and final velocities is,
$\Rightarrow {v^2} = {u^2} + 2as$
Let the distances covered by the stones be ${h_1}$ and ${h_2}$ .
For the first stone,
$\Rightarrow {v^2} = {u_1}^2 + 2a{h_1}$
To get ${h_1}$, substitute the values $v = 0$, in the equation above
$\Rightarrow 0 - {u_1}^2 = 2( - g){h_1}$
Therefore, we get the distance as,
$\Rightarrow {h_1} = \dfrac{{{u_1}^2}}{{2g}}$
For the second stone,
$\Rightarrow {v^2} = {u_2}^2 + 2a{h_2}$
To get ${h_2}$, substitute the values $v = 0$, in the equation above
$\Rightarrow 0 - {u_2}^2 = 2( - g){h_2}$
Therefore, we get the distance as,
$\Rightarrow {h_2} = \dfrac{{{u_2}^2}}{{2g}}$
Now, to find the ratio of maximum heights reached by the two stones,
We need to divide the height of the first stone to that of the second stone.
So, we get,
$\Rightarrow \dfrac{{{h_1}}}{{{h_2}}} = \dfrac{{\dfrac{{{u_1}^2}}{{2g}}}}{{\dfrac{{{u_2}^2}}{{2g}}}}$
After simplification and cancelling the common terms we get
$\Rightarrow \dfrac{{{h_1}}}{{{h_2}}} = \dfrac{{{u_1}^2}}{{{u_2}^2}}$
Hence, the ratio of maximum heights reached by the stones is equal to the ratio of the initial velocities of the two stones.

Note:
When two stones are thrown vertically upwards, the acceleration due to gravity acting on them will be negative, because it is opposite to the direction of gravity which is downward. Thus, when considering something thrown upwards, we need to apply the sign as negative.