We earliar studied about Projectile motion when ball is thrown up from ground at an angle.Now in this page , we will see Projectile Motion on a Inclined Plane
Projectile Motion on a Inclined Plane is the motion when a ball or object is thrown at an angle up the inclined plane or down the plane.
Lets check how to obtain various things like time of flight, range and maximum in each case
Motion Up The inclined plane
Let the particle be thrown up at an angle $\theta$ from the horizontal on the inclined plane of angle $\alpha$.
We can obtained the equation of motion using both X-Y axis and X' -Y' plane as shown in the figure. X' -Y' plane will be more convienient
In X' -Y'
we have
$a_{x'}=-g \sin \alpha$
$a_{y'}=-g \cos \alpha$
$u_{x'}=v_0 cos( \theta - \alpha)$
$u_{y'}=v_0 sin( \theta - \alpha)$
Now equation of motion along both the x' and y' will be
$x'= v_0 cos( \theta - \alpha) t - \frac {1}{2}g sin \alpha t^2$
$y'= v_0 sin( \theta - \alpha) t - \frac {1}{2}g cos \alpha t^2$
Now for Time of Flight t=T, we have y'=0,so
$0= v_0 sin( \theta - \alpha) t - \frac {1}{2}g cos \alpha t^2$
or $T = \frac {2 v_0 sin (\theta - \alpha)}{g cos \alpha}$
So range along the inclined plane will be given by
$x'= v_0 cos( \theta - \alpha) \frac {2 v_0 sin (\theta - \alpha)}{g cos \alpha} - \frac {1}{2}g sin \alpha (\frac {2 v_0 sin (\theta - \alpha)}{g cos \alpha})^2$
$x'=\frac {2v_0^2 sin (\theta - \alpha)}{g cos \alpha}[cos( \theta - \alpha) - \frac {sin \alpha sin (\theta - \alpha)}{cos \alpha}]$
$=\frac {2v_0^2 sin (\theta - \alpha)}{g cos^2 \alpha}[cos \alpha cos( \theta - \alpha) - sin \alpha sin (\theta - \alpha)]$
Now cos(A+B)=(cosAcosB−sinAsinB)
Therefore
$=\frac {2v_0^2 cos \theta sin (\theta - \alpha)}{g cos^2 \alpha}$
Now Applying formula 2 Cos A Sin B=Sin (A+B) - Sin (A-B)
we have
$R=\frac {v_0^2[sin (2\theta -\alpha) - sin \alpha]}{g cos^2 \alpha}$
In X-Y
we have
$a_{x}=0$
$a_{y}=-g$
$u_{x}=v_0 cos( \theta)$
$u_{y}=v_0 sin( \theta)$
Now equation of motion along both the x and y will be
$x= v_0 cos( \theta) t $
$y= v_0 sin( \theta) t - \frac {1}{2}g t^2$
Now when the projectile hit the inclined plane
$x=Rcos \alpha$
$y=R sin \alpha$
Therefore
$Rcos \alpha= v_0 cos( \theta) t $
$R sin \alpha= v_0 sin( \theta) t - \frac {1}{2}g t^2$
Eliminating t between these two
$R= \frac {2v_0^2 cos^2 \theta}{g cos \alpha}[tan \theta - tan \alpha]$
we can deduce this to same formula
$R=\frac {v_0^2[sin (2\theta -\alpha) - sin \alpha]}{g cos^2 \alpha}$
Now for Maximum Range, we should $sin (2\theta -\alpha)$ as maximum i.e $sin (2\theta -\alpha)=1$
So Maximum Range will be given as
$R=\frac {v_0^2[sin (2\theta -\alpha) - sin \alpha]}{g cos^2 \alpha}= \frac {v_0^2[1- sin \alpha]}{g cos^2 \alpha}= \frac {v_0^2}{g(1+ sin \alpha)}$
Motion down The inclined plane
Let the particle be thrown down at an angle $\theta$ from the horizontal on the inclined plane of angle $\alpha$.
We can obtained the equation of motion using both X-Y axis and X' -Y' plane as shown in the figure. X' -Y' plane will be more convienient
In X' -Y'
we have
$a_{x'}=-g \sin \alpha$
$a_{y'}=-g \cos \alpha$
$u_{x'}=v_0 cos( \theta + \alpha)$
$u_{y'}=v_0 sin( \theta + \alpha)$
Now equation of motion along both the x' and y' will be
$x'= v_0 cos( \theta + \alpha) t + \frac {1}{2}g sin \alpha t^2$
$y'= v_0 sin( \theta + \alpha) t - \frac {1}{2}g cos \alpha t^2$
Now for Time of Flight t=T, we have y'=0,so
$0= v_0 sin( \theta + \alpha) t - \frac {1}{2}g cos \alpha t^2$
or $T = \frac {2 v_0 sin (\theta + \alpha)}{g cos \alpha}$
So range along the inclined plane will be given by
$R= v_0 cos( \theta + \alpha) T + \frac {1}{2}g sin \alpha T^2$
$=\frac {v_0^2[sin (2\theta +\alpha) + sin \alpha]}{g cos^2 \alpha}$
Now for Maximum Range, we should $sin (2\theta +\alpha)$ as maximum i.e $sin (2\theta +\alpha)=1$
So Maximum Range will be given as
$R=\frac {v_0^2[sin (2\theta +\alpha) + sin \alpha]}{g cos^2 \alpha}= \frac {v_0^2[1+ sin \alpha]}{g cos^2 \alpha}= \frac {v_0^2}{g(1- sin \alpha)}$
Solved Examples
Question 1
A football is kicked on a inclined plane of angle 30° at speed 18m/s and angle 60° from the horizontal. Find
1. How much later does it hit the inclined plane?
2.How far from its initial does it hit the incline place? Solution
Here $v_0=18$ m/s , $\theta =60$ ° and $\alpha =30$°
Time of Flight is given by
$T = \frac {2 v_0 sin (\theta - \alpha)}{g cos \alpha}=\frac {2 \times 18 \times .5}{10 \times .866}=1.16$ sec
Range is given by
$R=\frac {v_0^2[sin (2\theta -\alpha) - sin \alpha]}{g cos^2 \alpha}$=\frac {18 \times 18 \times .5}{10 \times .866 \times .866}=21.60$ m
