Question

A projectile is fired with a velocity v0 at an angle of 90° with an inclined plane (having angle of inclination θ with the horizontal). Vertical displacement of projectile when it strikes the incline for first time is

Answer 1

Vertical displacement of projectile when it strikes the incline for first time is

$\boxed{\frac{v_0^2\tan^2\theta}{g}}$

Explanation:

From projectile motion on an inclined plane, we know that if a projectile is fired with a velocity $u$ from a plane inclined with angle $\alpha$ and if the angle of projection $\theta$ w.r.t. the inclined plane then

The range of the projectile

$R=\frac{u^2}{g\cos^2\alpha}[\sin (2\theta+\alpha)-\sin\alpha]$

Here,

Given that

$u=v_0$

$\theta=90^\circ$

$\alpha=\theta$

Therefore, range

$R=\frac{v_0^2}{g\cos^2\theta}[\sin (2\times 90^\circ+\theta)-\sin\theta]$

$\implies R=\frac{v_0^2}{g\cos^2\theta}[\sin (180^\circ+\theta)-\sin\theta]$

$\implies R=\frac{v_0^2}{g\cos^2\theta}[-2\sin\theta]$

$\implies R=-\frac{v_0^2\sin\theta}{g\cos^2\theta}$

Negative sign indicates that the projectile falls down the inclined plane

If the vertical displacement is d then

$\sin\theta=\frac{d}{R}$

$\implies d=R\sin\theta$

$\implies d=\frac{v_0^2\sin\theta}{g\cos^2\theta}\times\sin\theta$

$\implies d=\frac{v_0^2\tan^2\theta}{g}$

Hope this answer is helpful.

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