Question

A stone is projected from a horizontal plane. it attains maximum height 'H' and strikes the stationary smooth wall and falls on the ground vertically below the maximum height. Assume the collision to be elastic, the height of the point on the wall where the stone will strike is :-


a)H/2
b) H/4
c) 3H/4
d) None of these
​

Answer 1

So the stone follows the path AEBCD as in the attachment.

Since the collision is elastic, if there was no wall then the stone would follow the parabolic path AEBCD', and also the path CD' is the exact reflection of CD wrt the ray GC (or the wall).

That implies GD = GD'   ...(1)

Now,

• AD' = R, range of the parabolic path

• BD = H, maximum height

Then B is the vertex of the parabola, D is the midpoint of AD', and so,

$\sf{\longrightarrow DD'=\dfrac{R}{2}}$

$\sf{\longrightarrow GD+GD'=\dfrac{R}{2}}$

According to (1),

$\sf{\Longrightarrow GD=GD'=\dfrac{R}{4}}$

ED is the reflection of CD wrt to BD. Then,

$\sf{\longrightarrow GC=FE=h}$

$\sf{\longrightarrow GD=FD=\dfrac{R}{4}}$

Also,

$\sf{\longrightarrow AF=AD'-DD'-FD}$

$\sf{\longrightarrow AF=R-\dfrac{R}{2}-\dfrac{R}{4}}$

$\sf{\longrightarrow AF=\dfrac{R}{4}}$

Thus, consider the stone reaches the point E after a time t of projection. The vertical displacement of the stone at this time (FE) is equal to the height of the point on the wall where the stone will strike (GC).

Let u be initial velocity, θ be angle of projection. Then horizontal displacement at this time t will be,

$\sf{\longrightarrow AF=u\cos\theta\,t}$

$\sf{\longrightarrow\dfrac{R}{4}=u\cos\theta\,t}$

$\sf{\longrightarrow t=\dfrac{R}{4u\cos\theta}\quad\quad\dots(2)}$

And the vertical displacement will be,

$\sf{\longrightarrow FE=u\sin\theta\,t-\dfrac{1}{2}\,gt^2}$

$\sf{\longrightarrow h=u\sin\theta\,t-\dfrac{1}{2}\,gt^2}$

Putting value of t from (2),

$\sf{\longrightarrow h=u\sin\theta\cdot\dfrac{R}{4u\cos\theta}-\dfrac{g}{2}\cdot\dfrac{R^2}{16u^2\cos^2\theta}}$

$\sf{\longrightarrow h=\dfrac{R\sin\theta}{4\cos\theta}-\dfrac{gR^2}{32u^2\cos^2\theta}}$

Since the horizontal range, $\sf{R=\dfrac{u^2\sin(2\theta)}{g}=\dfrac{2u^2\sin\theta\cos\theta}{g},}$

$\sf{\longrightarrow h=\dfrac{2u^2\sin\theta\cos\theta}{g}\cdot\dfrac{\sin\theta}{4\cos\theta}-\dfrac{g}{32u^2\cos^2\theta}\cdot\dfrac{4u^4\sin^2\theta\cos^2\theta}{g^2}}$

$\sf{\longrightarrow h=\dfrac{u^2\sin^2\theta}{2g}-\dfrac{u^2\sin^2\theta}{8g}}$

$\sf{\longrightarrow h=\dfrac{u^2\sin^2\theta}{2g}-\dfrac{u^2\sin^2\theta}{2g}\cdot\dfrac{1}{4}}$

$\sf{\longrightarrow h=\dfrac{u^2\sin^2\theta}{2g}\left[1-\dfrac{1}{4}\right]}$

$\sf{\longrightarrow h=\dfrac{u^2\sin^2\theta}{2g}\cdot\dfrac{3}{4}}$

Since $\sf{\dfrac{u^2\sin^2\theta}{2g}=H,}$ the maximum height,

$\sf{\longrightarrow\underline{\underline{h=\dfrac{3H}{4}}}}$

Hence (c) is the answer.

Answer 2

Question:

A stone is projected from a horizontal plane. it attains maximum height 'H' and strikes the stationary smooth wall and falls on the ground vertically below the maximum height. Assume the collision to be elastic, the height of the point on the wall where the stone will strike is :-

• H/2
• H/4
• 3H/4
• None of these

Given:

Max height= H

$\sf H= \frac{u^2sin^2 \theta} {2g}$

To find:

The height at which the ball strikes the wall. Let it be h.

Let the time taken to strike is t

Let the initial velocity of the ball be u.

The horizontal component of velocity is $\sf ucos\theta$

And, the vertical component will be $\sf usin\theta$

Solution:

Since, the collision of ball with the wall is elastic, the energy of the ball will be conserved and so its velocity will not change. After collision the ball will travel the same distance as it would have been travelled if there was no wall in between. Only the direction of Velocity will change, not the magnitude.

So , QS=SR=QR/2 - - - [i]

For an instánce, let's assume that, there's no wall between the trajectory. It would have fallen on ground at point R. And the range would have been PR.

$\sf PR = \frac{u^2cos \theta sin \theta} {g}$

QT=H is its max height. It will reach its max height in the middle of the trajectory and thus it will symmetrically divide the trajectory into two equal halves. (see the attachment)

$\sf PQ=QR=\frac{PR}{2}- - - [ii]$

From [i], we have,

QS=SR=QR/2

Putting the value of QR from [ii], we obtain.

$\sf QS=SR= \frac{\frac{PR}{2}}{2}\\\\ \sf QS=SR= \frac{PR}{4}$

Hence, The distance travelled by the ball after collision is $\sf \frac{PR}{4}$

Now, the distance from P to S i.e where it strikes the wall is $\sf PQ+QS= \frac{PR}{2}+\frac{PR}{4}=\frac{3PR}{4}$

Now, using formula of range, we have,

$\sf \frac{3PR}{4}=u~cos \theta ~t \\\\ \sf t= \frac{3PR}{4ucos \theta} - - - [iii]$

Using $\boxed{s=ut- \frac{1}{2}gt^2}$ for displacement in vertical direction. We have,

$h= usin \theta ~t- \frac{1}{2}gt^2 - - - [iv]$

Putting the value of t from [iii] in [iv], we have,

$h = usin \theta \times \frac{3PR}{4u cos \theta} - \frac{g}{2} \times \bigg(\frac{3PR}{4u cos \theta} \bigg)^2$

Putting, $PR = \frac{u^2cos \theta sin \theta} {g}$

We have,

$h= \frac{sin \theta \times 3u^2 sin \theta} {2g}- \frac{9u^2sin^2 \theta}{8g}$

Taking $\frac{u^2sin^2 \theta} {2g}$ common:

$h= \frac{u^2sin^2 \theta}{2g} \bigg(3- \frac{9}{4}\bigg) \\\\ h= \frac{u^2sin^2 \theta}{2g} \bigg( \frac{3}{4} \bigg)$

We know that $\frac{u^2sin^2 \theta} {2g} = H$

So, $h = \frac{3}{4}H$

Hence, option c is correct.