Question

A stone is projected from a horizontal plane. It attains maximum height ${}^{\prime }{H}^{\prime }$ & strikes a stationary smooth wall & falls on the ground vertically below the maximum height. Assume the collision to be elastic, the height of the point on the wall where ball will strike is

(Elastic collision means that after colliding on a plane the perpendicular component of the velocity before collision gets reversed after the collision. Here $v_x$ is normal to plane)

• A. $\frac{H}{2}$
• B. $\frac{H}{4}$
• C. $\frac{3H}{4}$
• D. None of these

Answer 1

The correct option is C

$\frac{3H}{4}$

The ball will cover the same distance as only velocity's direction is changed magnitude is the same. So the same distance will be covered in opposite direction.

Now from the figure it's clear,

$2x$ = $\frac{R}{2}$

$\Rightarrow$ $x$ = $\frac{R}{4}$

$\Rightarrow$

So it means I have to find the height of the ball when it has travelled $\frac{3R}{4}$ along the x-axis.

Method I

Assume initial speed of projection is $u$ and angle $\theta$

$\Rightarrow$ $ucos\theta t$ = $\frac{3R}{4}$

$\Rightarrow$ $t$ = $\frac{3R}{4ucos\theta }$ ..........(1)

$h$ = $usin\theta t-\frac{1}{2}g{t}^2$ ..........(2)

substituting (I) in (II) we get

$h$ = $\frac{usin\theta \times 3R}{4ucos\theta }-\frac{1}{2}g\frac{9R^2}{16u^{2}co{s}^2\theta }$

we know $R$ in this case is $\frac{2u^{2}sin\theta cos\theta }{g}$

$\Rightarrow$ $h$ = $\frac{usin\theta X3X2u^{2}sin\theta cos\theta }{24ucos\theta g}-\frac{1}{2}g\frac{9\times {4u^4}^{2}si{n}^2\theta co{s}^2\theta }{g^{2}g\times {16}^4{u}^{2}co{s}^2\theta }$

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

$h$ = $\frac{u^{2}si{n}^2\theta }{2g}[3-\frac{9}{4}]$

we know $\frac{u^{2}si{n}^2\theta }{2g}$ = $H$

$\Rightarrow$ $h$ = $\frac{3}{4}H$.

Method II

Equation of trajectory

$y$ = $xtan\theta -\frac{t^{2}tan\theta }{R}$

$y$ = $h$

$x$ = $\frac{3R}{4}$

$\Rightarrow$ $h=\frac{3R}{4}tan\theta -\frac{9R^{2}R}{16}\frac{tan\theta }{R}$

$h=\frac{3\times 2\times u^{2}sin\theta cos\theta sin\theta }{24\times gcos\theta }-\frac{9\times 2u^{2}sin\theta cos\theta sin\theta }{816\times u^2\times gcos\theta }$

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

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

$h$ = $\frac{3}{4}H$.