**Question:**

A projectile is fired with a speed $u$ at an angle ${ }^{\theta}$ above a horizontal field. The coefficient of restitution of collision between the projectile and the field is e. How far from the starting point, does the projectile makes its second collision with the field?

**Solution:**

$\mathrm{e}=$ co-efficient of restitution

$$

u_{x}=u \cos \theta

$$

After collision with ground, $u_{y}{ }^{x}=e u \sin \theta$

$u_{x}{ }^{x}=u_{x}$

$u=\sqrt{\left(u x^{2}\right)^{2}+\left(u y^{\circ}\right)^{2}}=\sqrt{(e u \sin \theta)^{2}+(u \cos \theta)^{2}}$

$\alpha=\tan ^{-1}\left(\frac{u_{y}}{u_{x}^{\prime}}\right)=\tan ^{-1}(e \tan \theta)$

$y=x \tan \alpha-\frac{g x^{2} \sec ^{2} \alpha}{2 u^{2}}$ is the equation of trajectory.

$\Rightarrow 0=x(e \tan \theta)-\frac{g x^{2}\left(1+e^{2} \tan ^{2} \theta\right)}{2 u^{2}\left(\cos ^{2} \theta+e^{2} \sin ^{2} \theta\right)}$

$\left\{\mathrm{y}=0\right.$ at second projectile after it falls down from $1^{\text {st }}$ projectile $\}$

$\Rightarrow x=\frac{2 e u^{2} \tan \theta \cdot \cos ^{2} \theta}{g}=\frac{e u^{2} \sin 2 \theta}{g}$

From starting point, it falls after distance.

$x^{\prime \prime}=x+\frac{e u^{2} \sin 2 \theta}{g}=\frac{u^{2} \sin 2 \theta}{g}+\frac{e u^{2} \sin 2 \theta}{g}$

$=\frac{u^{2} \sin 2 \theta}{g}(1+e)$

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.