Question

A particle is projected with speed $V_o$ at an angle $\theta$ to the horizontal on an inclined surface making an angle $\varphi$ $(\varphi <\theta )$ to the horizontal. Find an expression for its range along the inclined surface.

Answer 1

$u_X=v_0\cos \left(\theta -\varphi \right)$

$u_Y=v_0\sin \left(\theta -\varphi \right)$

$a_X=-g\sin \varphi$

$a_Y=-g\cos \varphi$

$R=u_{X}t-\frac{1}{2}g\sin \varphi t^2$

$v_Y-0$

$u_Y+a_{Y}t=0$

$v_0\sin \left(\theta -\varphi \right)$-$g\cos \varphi t=0$

$t=\frac{v_0\sin (\theta -\varphi )}{g\cos \varphi }$

Time of flight =$T$=$\frac{2v_0\sin (\theta -\varphi )}{g\cos \varphi }$

$R=u_{X}T-\frac{1}{2}g\sin \varphi T^2$

$=T\left(v_0\cos \right(\theta -\varphi )-\frac{1}{2}g\sin \varphi T)$

$=T\left(v_0\cos \right(\theta -\varphi )-\frac{g\sin \varphi }{2}\times \frac{2v_0\sin (\theta -\varphi )}{g\cos \varphi })$

$=T{v}_0\left(\cos \right(\theta -\varphi )-\frac{\sin \varphi \sin (\theta -\varphi )}{\cos \varphi })$

$=v_{0}T(\frac{\cos (\theta -\varphi )\cos \varphi -\sin \varphi \sin (\theta -\varphi )}{\cos \varphi }$

$=v_{0}T\frac{\cos \theta }{\cos \varphi }$

$=v_0\frac{\cos \theta }{\cos \varphi }\times \frac{2v_0\sin (\theta -\varphi )}{g\cos \varphi }$

$R=\frac{2v_0^2\sin (\theta -\varphi )\cos \theta }{g{\cos }^2\varphi }$