Question

A body, thrown upward with some velocity reaches the maximum height of $50 m$. Another body with double the mass thrown up with double the initial velocity will reach a maximum height of
A. $200 m$
B. $16 m$
C. $80 m$
D. $40 m$

Answer 1

We can use a kinematic equation relating final velocity, initial velocity, acceleration and time to determine the maximum height attained by the body. We know that at maximum height, the velocity of the body becomes zero.

Formula used:
${v^2} = {u^2} + 2as$
Here, v is the final velocity, u is the initial velocity, a is the acceleration and t is the time.

Complete step by step answer:
To solve this question, we can use a kinematic equation relating final velocity, initial velocity, acceleration and time taken.
${v^2} = {u^2} + 2as$
We can use the above equation in the vertical direction as follows,
${v^2} = {u^2} - 2gh$
Here, g is acceleration due to gravity and h is the maximum height attained by the body.

We know that at maximum height, the final velocity of the body becomes zero. Therefore, the above equation becomes,
${u^2} = 2gh$
We can express the above equation for both bodies as follows,
$u_1^2 = 2g{h_1}$ …… (1)
And,
$u_2^2 = 2g{h_2}$ …… (2)
Divide equation (2) by equation (1).
$\dfrac{{u_2^2}}{{u_1^2}} = \dfrac{{2g{h_2}}}{{2g{h_1}}}$
$\Rightarrow \dfrac{{u_2^2}}{{u_1^2}} = \dfrac{{{h_2}}}{{{h_1}}}$
$\Rightarrow {h_2} = {h_1}\dfrac{{u_2^2}}{{u_1^2}}$

We substitute 50 m for ${h_1}$, $2{u_1}$ for ${u_2}$ in the above equation.
${h_2} = \left( {50\,m} \right)\dfrac{{{{\left( {2{u_1}} \right)}^2}}}{{u_1^2}}$
$\Rightarrow {h_2} = \left( {50\,m} \right)\left( 4 \right)$
$\therefore {h_2} = 200\,m$

So, the correct answer is option (A).

Note: Students can also solve this question using the formula for maximum height attained by the projectile, ${H_{\max }} = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$, where, $\theta$ is the angle made by the projectile with horizontal. By substituting $\theta = 90^\circ$ we can get the same equation we have found out in the solution. Students should remember that the acceleration due to gravity for the body moving upward is taken as negative.