Question

Two shells are fired from a cannon successively with speed $u$, each at angles of projection $\alpha$ and $\beta$ respectively. If the time interval between the firing of shots is $t$ and they collide in mid-air after a time $T$ from the firing of the first shot. Then

• A. $Tcos\alpha =(T-t)cos\beta$
• B. $\alpha >\beta$
• C. $(T-t)cos\alpha =tcos\beta$
• D. $\left(usin\alpha \right)T-\frac{1}{2}{gT}^2=\left(usin\beta \right)(T-t)-\frac{1}{2}g(T-t{)}^2$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $Tcos\alpha =(T-t)cos\beta$
B $\alpha >\beta$
D $\left(usin\alpha \right)T-\frac{1}{2}{gT}^2=\left(usin\beta \right)(T-t)-\frac{1}{2}g(T-t{)}^2$

A $\to$ angle $\alpha \to$ fired first

Total time of flight before collision $=T$
$u_n=u\cos \alpha S_n=u\cos \alpha T$
$u_v=u\sin \alpha a=-g{S}_v=u\sin \alpha T-\frac{1}{2}g{T}^2$

B $\to$ angle $\beta$ ;

total time of flight $=T-t$

$u_n=u\cos \beta S_n^{\prime }=u\cos \beta (T-t)$
$u_v=u\sin \beta a=-g{S}_v^{\prime }=u\sin \beta (T-t)-\frac{1}{2}g(T-t{)}^2$

Given that they collide in mid air:

$S_n=S_n^{\prime }$ & $S_v=S_v^{\prime }\Rightarrow u\cos \alpha T=u\cos \beta (T-t)\Rightarrow T\cos \alpha =(T-t)\cos \beta$

option A is correct

$\because T>T-t\Rightarrow \cos \alpha <\cos \beta \Rightarrow \alpha >\beta$

option B is correct

$u\sin \alpha T-\frac{1}{2}g{T}^2=u\sin \beta (T-t)-\frac{1}{2}g(T-t{)}^2$

option D is correct