Question

Two shells are fired from a canon with a speed $u$ each, at angles of $\alpha$ and $\beta$, respectively, to the horizontal. The time interval between the shots is $t$. They collide in mid-air after time $T$ from the first shot. Which of the following conditions must be satisfied?

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

Answer 1

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

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

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 $=T-t$
$u_n=u\cos \beta S_n^{\prime }=u\cos \beta (T-t)$
$u_v=u\sin \beta a=-g{S}_v^{\prime }=u\sin \beta (T-t)-\frac{1}{2}g(T-t{)}^2$
$S_n=S_n^{\prime }$ & $S_v=S_v^{\prime }\Rightarrow u\cos \alpha T=u\cos \beta (T-t)\Rightarrow T\cos \alpha =(T-t)\cos \beta$
$\because T>T-t\Rightarrow \cos \alpha <\cos \beta \Rightarrow \alpha >\beta$
$u\sin \alpha T-\frac{1}{2}g{T}^2=u\sin \beta (T-t)-\frac{1}{2}g(T-t{)}^2$
Hence A, B, D