Question

The direction of a projectile at a certain instant is inclined at an angle $\alpha$ to the horizontal; after $t$ second, it is inclined at an angle $\beta$. Prove that the horizontal component of the
velocity of the projectile is $\frac{gt}{tan\alpha -tan\beta }$

Answer 1

Let the projectile be at point $P$ at any instant with its velocity inclined at an angle $\alpha$ to the horizontal.

There for horizontal component $=u\cos \alpha$...........(i)

and vertical component $=u\sin \alpha$ ................(ii)

where $u$ is the velocity of the projectile at $P$

let after $t$ seconds the particle reaches point $Q$ with angle of inclination $\beta$ to the horizontal and velocity $v$

Resolving velocity at $Q$

Horizontal component =$=v\cos \beta$..................(iii)

and vertical component =$v\sin \beta$............(iv)

Therefore horizontal component remains same

$\therefore u\sin \alpha =v\cos \beta$

and for the vertical motion from poit $P$ to $Q$,

$u=u\sin \alpha ;v=v\sin \beta ;a=-g;t=t$

using $v=u+at$

$v\sin \beta =u\sin \alpha -gt$

substituting for $v$ from Eq.(vi in Eq(v),

$u\cos \alpha =\left(\frac{u\sin \alpha -gt}{\sin \beta }\right)\cos \beta =(u\sin \alpha -gt)\cos \beta$

$u\sin \alpha .\cot \beta -u\cos \alpha =gt.\cot \beta$

$u\cos \alpha [\tan \alpha .\cot \beta -1]=gt.\cot \beta$

$u\cos \alpha =\frac{gt\cos \beta }{\tan \alpha .cot\beta -1}=\frac{gt}{\tan \alpha -\tan \beta }$ [Multiplying Nr. and Dr. by $\tan \beta$]

Therefore , horizontal component,

$u\cos \alpha =\frac{gt}{\tan \alpha -\tan \beta }$