# 5.7 Projectile motion on an incline (application)  (Page 2/2)

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## Angle of projection

Problem : A particle is projected from the foot of an incline of angle “30°” at a certain velocity so that it strikes the incline normally. Find the angle of projection (θ ) as measured from the horizontal.

Solution : Here, projectile hits the incline normally. It means that component of velocity along the incline is zero. We should remember that the motion along the incline is not uniform motion, but a decelerated motion. In order to take advantage of the fact that final component velocity along incline is zero, we consider motion in a coordinate system along the incline and along a direction perpendicular to it.

${v}_{x}={u}_{x}+{a}_{x}T$

$⇒0=u\mathrm{cos}\left(\theta -{30}^{0}\right)-g\mathrm{sin}{30}^{0}T$

$T=\frac{u\mathrm{cos}\left(\theta -{30}^{0}\right)}{g\mathrm{sin}{30}^{0}}$

Now, time of flight is also given by the formulae :

$⇒T=\frac{2u\mathrm{sin}\left(\theta -{30}^{0}\right)}{g\mathrm{cos}{30}^{0}}$

Equating two expressions for time of flight, we have :

$⇒\frac{2u\mathrm{sin}\left(\theta -{30}^{0}\right)}{g\mathrm{cos}{30}^{0}}=\frac{u\mathrm{cos}\left(\theta -{30}^{0}\right)}{g\mathrm{sin}{30}^{0}}$

$⇒\mathrm{tan}\left(\theta -{30}^{0}\right)=\frac{1}{2\mathrm{tan}{30}^{0}}=\frac{\sqrt{3}}{2}$

$⇒\theta ={30}^{0}+{\mathrm{tan}}^{-1}\left(\frac{\sqrt{3}}{2}\right)$

## Final speed of the projectile

Problem : A ball is projected on an incline of 30° from its base with a speed 20 m/s, making an angle 60° from the horizontal. Find the speed with which the ball hits the incline.

Solution : We analyze this problem in the coordinates along the incline (x-axis) and in the direction perpendicular to the incline (y-axis). In order to find the speed at the end of flight, we need to find the component velocities in “x” and “y” directions.

The velocity in y-direction can be determined making use of the fact that a ball under constant acceleration like gravity returns to the ground with the same speed, but in opposite direction. The component of velocity in y-direction at the end of the journey, therefore, is :

${v}_{y}=-{u}_{y}=-20\mathrm{sin}{30}^{0}=-20X\frac{1}{2}=-10\phantom{\rule{1em}{0ex}}m/s$

Now, we should attempt to find the component of velocity in x-direction. We should, however, recall that motion in x-direction is not a uniform motion, but has deceleration of “-gsinα”. Using equation of motion in x-direction, we have :

${v}_{x}={u}_{x}+{a}_{x}T$

Putting values,

${v}_{x}=u\mathrm{cos}{30}^{0}-g\mathrm{sin}{30}^{0}T=20X\frac{\sqrt{3}}{2}-10X\frac{1}{2}XT=10\sqrt{3}-5T$

Clearly, we need to know time of flight to know the component of velocity in x-direction. The time of flight is given by :

$T=\frac{2{u}_{y}}{{a}_{y}}=\frac{2u\mathrm{sin}{30}^{0}}{g\mathrm{cos}{30}^{0}}=\frac{2X20}{\sqrt{3}X10}=\frac{4}{\sqrt{3}}$

Hence, component of velocity in x-direction is :

$⇒vx=10\sqrt{3}-5T=10\sqrt{3}-5X\frac{4}{\sqrt{3}}$

$vx=\frac{30-20}{\sqrt{3}}=\frac{10}{\sqrt{3}}$

The speed of the projectile is equal to resultant of components :

$⇒v=\sqrt{\left\{{\left(-10\right)}^{2}+{\left(\frac{10}{\sqrt{3}}\right)}^{2}\right\}}$

$v=\sqrt{\left(\frac{400}{3}\right)}=\frac{20}{\sqrt{3}}\phantom{\rule{1em}{0ex}}m/s$

## Elastic collision with the incline

Problem : A ball falls through a height “H” and impacts an incline elastically. Find the time of flight between first and second impact of the projectile on the incline.

Solution : The ball falls vertically through a distance "H". We can get the value of initial speed by considering free fall of the ball before impacting incline.

$⇒0={u}^{2}-2gH$

$⇒u=\sqrt{\left(2gH\right)}$

In order to answer this question, we need to identify the velocity with which projectile rebounds. Since impact is considered elastic, the projectile is rebounded without any loss of speed. The projectile is rebounded such that angle of incidence i.e. the angle with the normal is equal to angle of reflection.

The components of velocity along the incline and perpendicular to it are shown in the figure. The motion of ball, thereafter, is same as that of a projectile over an incline. Here, we shall analyze motion in y-direction (normal to the incline) to find the time of flight. We note that the net displacement between two strikes is zero in y – direction.

Applying equation of motion

$y={u}_{y}T+\frac{1}{2}{a}_{y}{T}^{2}$

$⇒0=u\mathrm{cos}\alpha T-\frac{1}{2}g\mathrm{cos}\alpha {T}^{2}$

$T=0$

Or

$T=\frac{2u\mathrm{cos}\alpha }{g\mathrm{cos}\alpha }=\frac{2u}{g}$

Putting value of initial speed in the equation of time of flight, we have :

$⇒T=\sqrt{\left(\frac{8H}{g}\right)}$

It should be noted here that we can find the time of flight also by using standard formula of time of flight for projectile motion down the incline. The time of flight for projection down an incline is given as :

$T=\frac{2u\mathrm{sin}\left(\theta +\alpha \right)}{g\mathrm{cos}\alpha }$

We need to be careful while appropriating angles in the above expression. It may be recalled that all angles are measured from the horizontal. We redraw the figure to denote the value of angle of projection “θ” from the horizon.

$⇒T=\frac{2u\mathrm{sin}\left({90}^{0}-2\alpha +\alpha \right)}{g\mathrm{cos}\alpha }=\frac{2u\mathrm{sin}\left({90}^{0}-\alpha \right)}{g\mathrm{cos}\alpha }$

$⇒T=\frac{2u\mathrm{cos}\alpha }{g\mathrm{cos}\alpha }=\frac{2u}{g}=\sqrt{\left(\frac{8H}{g}\right)}$

## Projectile motion on two inclines

Problem : Two incline plane of angles 30° and 60° are placed touching each other at the base as shown in the figure. A projectile is projected at right angle with a speed of 10√3 m/s from point “P” and hits the other incline at point “Q” normally. Find the linear distance between PQ.

Solution : We notice here OPQ forms a right angle triangle at “O”. The linear distance, “PQ” is related as :

$P{Q}^{2}=O{P}^{2}+O{Q}^{2}$

In order to find “PQ”, we need to know “OP” and “OQ”. We can find “OP”, considering motion in y-direction.

$y=OP={u}_{y}T+\frac{1}{2}{a}_{y}{T}^{2}=0+\frac{1}{2}{a}_{y}{T}^{2}=\frac{1}{2}{a}_{y}{T}^{2}$

Similarly,

$x=OQ={u}_{x}T+\frac{1}{2}{a}_{x}{T}^{2}=10\sqrt{3}T+\frac{1}{2}{a}_{x}{T}^{2}$

In order to evaluate these two relations, we need to find components of accelerations and time of flight.

Considering first incline, we have :

${a}_{x}=-g\mathrm{cos}{30}^{0}=-10X\frac{\sqrt{3}}{2}=-5\sqrt{3}\phantom{\rule{1em}{0ex}}m/{s}^{2}$

${a}_{y}=-g\mathrm{sin}{30}^{0}=-10X\frac{1}{2}=-5\phantom{\rule{1em}{0ex}}m/{s}^{2}$

In order to find the time of flight, we can further use the fact that the component of velocity in x-direction i.e. along the second incline is zero. This, in turn, suggests that we can analyze motion in x-direction to obtain time of flight.

In x-direction,

${v}_{x}={u}_{x}+{a}_{x}T$

$⇒0={u}_{x}+{a}_{x}T$

$⇒T=-\frac{{u}_{x}}{{a}_{x}}$

Putting values in the equation and solving, we have :

$⇒T=-\frac{10\sqrt{3}}{-5\sqrt{3}}=2s$

Now, we can evaluate “x” and “y” displacements as :

$⇒y=OP=\frac{1}{2}{a}_{y}{T}^{2}=-\frac{1}{2}X5X{2}^{2}=-10\phantom{\rule{1em}{0ex}}m$

$⇒x=OQ=10\sqrt{3}x2+\frac{1}{2}\left(-5\sqrt{3}\right)X{2}^{2}$

$x=OQ=10\sqrt{3}$

Considering positive values, the linear distance, “PQ” is given as :

$PQ=\sqrt{\left(O{P}^{2}+O{Q}^{2}\right)}=\sqrt{\left\{{\left(10\right)}^{2}+{\left(10\sqrt{3}\right)}^{2}\right\}}=20m$

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