Question

A particle is thrown vertically upward with a speed of u from a tower of height H. the time taken by the particle to hit the ground is n times that taken by it to reach the highest point of its path. The relation between H, u and n is:
$\text{A}\text{. }gH={{(n-2)}^{2}}{{u}^{2}}$
$\text{B}\text{. }\dfrac{n{{u}^{2}}(n-2)}{2}=gH$
$\text{C}\text{. }gH=\dfrac{{{n}^{2}}{{u}^{2}}}{2}$
$\text{D}\text{. }(n-2)=\dfrac{gH}{{{u}^{2}}}$

Answer 1

Hint: If the initial velocity, the final velocity, and the acceleration of the body are known, one can find the time taken by the body to reach this final velocity from its initial velocity. Once the time is known, the displacement can be found.
Formula used:
$v=u+at$
$s=ut+\dfrac{1}{2}a{{t}^{2}}$.

Complete step-by-step solution:
The kinematic equations used in the solution:
$v=u+at$
$s=ut+\dfrac{1}{2}a{{t}^{2}}$.
u is initial velocity, v is final velocity, s is displacement, a is acceleration and t is time.
The particle is thrown vertically up with a speed of u from a height of an H.

It is said that the time taken by the particle is n times the time taken by it to reach the highest point in this flight. So, let us first calculate the time taken by the particle to reach the highest point. Let that be ‘t’. Since we know that initial velocity (u upwards) at time t=0 and the acceleration of the particle (g downwards), to find t we will use the kinematic equation $v=u+at$. If we consider the upward direction as positive direction and downward direction as negative then v=0, u=u, and a= -g.
Therefore, $0=u-gt\Rightarrow t=\dfrac{u}{g}$.
This implies that the time of flight of the particle (T) i.e. the time taken by the particle to reach the ground from the point of release will be $T=nt=\dfrac{nu}{g}$. When the particle reaches the ground, its displacement will be –H. Therefore, we can use the displacement equation,$s=ut+\dfrac{1}{2}a{{t}^{2}}$, where “s” = -H, u=u, t=T and a = -g. Substitute these values in the equation.
$\Rightarrow -H=uT+\dfrac{1}{2}(-g){{T}^{2}}$
Substitute the value of $T=\dfrac{nu}{g}$.
$\Rightarrow -H=u\left( \dfrac{nu}{g} \right)+\dfrac{1}{2}(-g){{\left( \dfrac{nu}{g} \right)}^{2}}$
Multiply both sides of the equation by (-1).
$\Rightarrow H=-u\left( \dfrac{nu}{g} \right)+\dfrac{1}{2}g{{\left( \dfrac{nu}{g} \right)}^{2}}$
Open up the brackets.
$\Rightarrow H=-\dfrac{n{{u}^{2}}}{g}+\dfrac{1}{2}g\left( \dfrac{{{n}^{2}}{{u}^{2}}}{{{g}^{2}}} \right)$
$\Rightarrow H=-\dfrac{n{{u}^{2}}}{g}+\dfrac{{{n}^{2}}{{u}^{2}}}{2g}$
Multiply both sides by g.
$\Rightarrow Hg=-n{{u}^{2}}+\dfrac{{{n}^{2}}{{u}^{2}}}{2}$
Take $n{{u}^{2}}$ as a factor common from the right-hand side of the equation.
$\Rightarrow Hg=n{{u}^{2}}\left( -1+\dfrac{n}{2} \right)=n{{u}^{2}}\left( \dfrac{n-2}{2} \right)$
Therefore, the relation between H, u and n is $Hg=n{{u}^{2}}\left( \dfrac{n-2}{2} \right)$. Hence, the correct option is (B).

Note: If the particle is thrown in such a way that its initial velocity makes an angle $\theta $ (other than $90^o$) with the horizontal axis. Then we can use the same method to find the relation between u, H, n with just replacing u with its vertical component i.e. $u\sin \theta $ in all the above equations. The final result will be $Hg=n{{\left( u\sin \theta \right)}^{2}}\left( \dfrac{n-2}{2} \right)$.