Question

A ball is projected with an initial speed $u$ at an angle $\theta$ from the horizontal ground. The coefficient of restitution between the ball and the ground is $e$. Find the position from the starting point when the ball will land on the ground for the $2^{nd}$ time.

• A. $\frac{e^2{u}^2\sin 2\theta }{g}$
• B. $\frac{(1-e^2)u^2\sin 2\theta }{g}$
• C. $\frac{(1-e)u^2\sin \theta \cos \theta }{g}$
• D. $\frac{(1+e)u^2\sin 2\theta }{g}$

Answer 1

The correct option is D $\frac{(1+e)u^2\sin 2\theta }{g}$

Motion of a particle projected at an angle $\theta$ with horizontal will be parabolic, and it is a symmetrical motion about $y-$axis.


Range of the particle when it strikes the ground for 1st time:
$R_1=\frac{u^2\sin 2\theta }{g}$

Just after collison, the $y-$component of velocity $v_y$ will change its magnitude $\&$ direction due to impulsive force from the ground in $+vey-$ direction.
$e=\frac{\text{speed of separation}}{\text{speed of approach}}$
$\Rightarrow e=\frac{v_y}{u\sin \theta }$
$\therefore v_y=eu\sin \theta$

Now, this will be a second projectile motion as shown in figure below.


New time of flight $T^{\prime }=\frac{2u_{\perp }}{g}=\frac{2eu\sin \theta }{g}$,
Hence horizontal distance(Range) travelled by ball when it lands on the ground for $2^{nd}$ time:
$R_2=u_x{T}^{\prime }$
$\Rightarrow R_2=u\cos \theta \times \frac{2eu\sin \theta }{g}$
$\therefore R_2=\frac{e{u}^2\sin 2\theta }{g}$

Hence, total distance travelled by the ball along the ground, when it lands for $2^{nd}$ time:
$X=R_1+R_2$
$\therefore X=\frac{u^2\sin 2\theta }{g}(1+e)$