5.1 Up the incline projectile motion

  Motion in Two Dimension

Consider a particle is projected with velocity $u$ making an angle $\theta$ with the inclined plane having inclination $\alpha$ with the horizontal.

5.1.1 Time of flight $(T)$

${\overrightarrow u _Y}=+u \sin \theta$
${\overrightarrow a _Y}=-g \cos \alpha$
${\overrightarrow s _{Y}}=0$
$t=T$

Therefore we can write kinematics equation which relates displacement and time as,
$${\overrightarrow s _Y} = {\overrightarrow u _Y}t + \frac{1}{2}{\overrightarrow a _Y}{t^2}$$$$0 = \left( {u\sin \theta } \right)T + \frac{1}{2}\left( { - g\cos \alpha } \right){T^2}$$$$T = \frac{{2u\sin \theta }}{{g\cos \alpha }}$$

5.1.2 Maximum height $\left( {{H_{\max }}} \right)$

At maximum height velocity perpendicular to the inclined plane is zero.

${\overrightarrow u _Y}=+u \sin \theta$
${\overrightarrow v _Y}=0$
${\overrightarrow a _Y}=-g \cos \alpha$
${\overrightarrow s _{Y}}=+H_{max}$

Therefore we can write kinematics equation which relates displacement and velocity as,
$${v^2} = {u^2} + 2{\overrightarrow a _Y}{\overrightarrow s _Y}$$$$0 = {u^2}{\sin ^2}\theta + 2\left( { - g\cos \alpha } \right)\left( { + {H_{\max }}} \right)$$$${H_{\max }} = \frac{{{u^2}{{\sin }^2}\theta }}{{2g\cos \alpha }}$$

5.1.3 Range $(R)$

In this case, the horizontal distance travelled by the projectile along the incline during time of flight is known as range.

${\overrightarrow u _X}=+u \cos \theta$
${\overrightarrow a _X}=-g \sin \alpha$
${\overrightarrow s _{X}}=+R$
$t = T = \frac{{2u\sin \theta }}{{g\cos \alpha }}$

Therefore we can write kinematics equation which relates displacement and time as,
$${\overrightarrow s _X} = {\overrightarrow u _X}t + \frac{1}{2}{\overrightarrow a _X}{t^2}$$$$ + R = \left( {u\cos \theta } \right)\left( {\frac{{2u\sin \theta }}{{g\cos \alpha }}} \right) + \frac{1}{2}\left( { - g\sin \alpha } \right){\left( {\frac{{2u\sin \theta }}{{g\cos \alpha }}} \right)^2}$$$$R = \frac{{2{u^2}\sin \theta \cos \theta }}{{g\cos \alpha }} - \frac{{4g{u^2}{{\sin }^2}\theta \sin \alpha }}{{2{g^2}{{\cos }^2}\alpha }}$$$$R = \frac{{2{u^2}\sin \theta \cos \theta }}{{g\cos \alpha }} - \frac{{2{u^2}{{\sin }^2}\theta \sin \alpha }}{{g{{\cos }^2}\alpha }}$$$$R = \frac{{2{u^2}\sin \theta }}{{g\cos \alpha }}\left( {\cos \theta - \frac{{\sin \theta \sin \alpha }}{{\cos \alpha }}} \right)$$$$R = \frac{{2{u^2}\sin \theta }}{{g\cos \alpha }}\left( {\frac{{\cos \theta \cos \alpha - \sin \theta \sin \alpha }}{{\cos \alpha }}} \right)$$$$R = \frac{{2{u^2}\sin \theta \cos \left( {\theta + \alpha } \right)}}{{g{{\cos }^2}\alpha }}$$
where,

$\alpha:$ angle which inclined plane makes with the horizontal
$\theta:$ angle of projection from the inclined plane