IIT JEE , AIEEE Forum - Physics , Mechanics

abhisek_mukh

how is a projectile on an inclined plane different from an ordinARY projectile and i cannot undrstand how to resolve the gravity as per need

mukundmadhav

Don't resolve gravity.. Obtain equation of motion by eliminating t from the displacement equations i x and y direction. Now run the equation of the incline with this. This will be a straight line

srujana

It is not different from the ordinary projectile motion,except that the vectors are resolved along the chosen axes

Let the angle of inclination of the incline be $\alpha$ and the angle of projection to be $\theta$ with respect to the ground.

v=velocity of the projectile

Consider two axes X,Y along and perpendicular to the incline

Along X axis

$v_x=vcos(\theta-\alpha)$

$a_x=-gsin\alpha$

Along Yaxis

$v_y=vsin(\theta-\alpha)$

$a_y=-gcos\alpha$

Write the equations of motion along X and Y axes

$Range(x)=v_xt+\frac{1}{2}a_xt^2$

$Height(h)=v_yt+\frac{1}{2}a_yt^2$

elessar_iitkgp

Good work @srujana. Slight correction, it should be ${v}_{x}={v}_{0}cos(\theta +\alpha)$ if $\theta$ is being measured form horizontal

Refer to the figure provided by srujana. Take origin at the point of projection, X axis downwards along the inclined plane, and Y axis perpendiculato to the plane upwards.

Then,Particle's initial position $\vec{{r}_{0}} = \vec{0}$

Initial velocity, $\vec{{v}_{0}}={v}_{0}(cos\theta\hat{i}+sin\theta\hat{j})$

And the acceleartion, $\vec{a}=g(sin\alpha \hat{i}-cos\alpha \hat{j})$

Then, using $\Delta\vec{r}=\vec{{v}_{0}}\Delta t + \frac{1}{2}\vec{a}(\Delta t)^2$ with the assumption that the particle is thrown at we get

$\vec{r}=x\hat{i}+y\hat{j}=({v}_{0}cos\theta+\frac{1}{2}gsin\alpha t^2)\hat{i}+({v}_{0}sin\theta-\frac{1}{2}gcos\alpha t^2)\hat{i}$

Giving, $x=({v}_{0}cos\theta t+\frac{1}{2}gsin\alpha t^2)$ and $y=({v}_{0}sin\theta t-\frac{1}{2}gcos\alpha t^2)$

Let at

, the particle hit the inclined plane again. Then $y(T)=0\Rightarrow T=\frac{2{v}_{0} sin\theta}{gcos\alpha}$

And at that instant the x coordinate gives the range of the particle on the inclined plane, i.e.,

$R=x(T) = \frac{2{v}_{0}^2sin\theta cos(\theta-\alpha)}{gcos^2\alpha}$

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