Question

A projectile is fired at an angle $\theta$ with the horizontal. Find the condition under which it lands perpendicular on an inclined plane inclination $\alpha$ as shown in figure.

• A.
  $\cos \alpha =\sin (\theta -\alpha )$
• B.
  $\tan \theta =\cot (\theta -\alpha )$
• C.
  $\cot (\theta -\alpha )=2\tan \alpha$
• D.
  $\sin \alpha =\cos (\theta -\alpha )$

Answer 1

The correct option is C

$\cot (\theta -\alpha )=2\tan \alpha$

Step 1: Draw the diagram showing the path of the projectile.
Step 2: Find time taken by the projectile to land on the incline.
Applying equation of motion perpendicular to the incline for $y=0$
$y=u_{y}t+\frac{1}{2}{a}_y{t}^2$
$0=V\sin (\theta -a)t+\frac{1}{2}(-g\cos \alpha )t^2$
$\Rightarrow t=0,\frac{2v\sin (\theta -\alpha )}{g\cos \alpha }$
Step 3: Find the condition foe which projectile lands perpendicular on the incline plane.

At the moment of striking the plane, as velocity is perpendicular to the inclined plane hence component of velocity along incline must be zero.

$v_x=u_x+a_{x}t$
$0=V\cos (\theta -\alpha )+(-g\sin \alpha )$
$\frac{2V\sin (\theta -\alpha )}{g\cos \alpha }$
$V\cos (\theta -\alpha )=\tan \alpha .2V\sin (\theta -\alpha )$
$\cot (\theta -\alpha )=2\tan \alpha$
Final answer: (D)