Question

From top of two towers of height $2\text{m}$ and $6\text{m}$, two bodies $A$ and $B$ are thrown simultaneously towards each other respectively. Body $A$ is thrown horizontally with a velocity of $8\text{m/s}$ while body $B$ is thrown downwards at an angle of ${45}^{\circ }$ to the horizontal at an initial velocity such that the two bodies collide during flight. The horizontal distance between the towers is $8\text{m}$. If the trajectories lie in a single plane, which of the following options are correct?

• A. The initial velocity $u_o$ of the body thrown at an angle of ${45}^{\circ }$ is $8\sqrt{2}\text{m/s}$
• B. The times of flight $t$ of the bodies before collision is $0.5\text{s}$
• C. Velocity of body $A$ ($v_A$) at the instant of collision is $9.4\text{m/s}$
• D. Velocity of body $B$ ($v_B$) at the instant of collision is $15.2\text{m/s}$

Answer 1

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

The correct options are
A The initial velocity $u_o$ of the body thrown at an angle of ${45}^{\circ }$ is $8\sqrt{2}\text{m/s}$
B The times of flight $t$ of the bodies before collision is $0.5\text{s}$
C Velocity of body $A$ ($v_A$) at the instant of collision is $9.4\text{m/s}$
D Velocity of body $B$ ($v_B$) at the instant of collision is $15.2\text{m/s}$


Let us analyze the relative velocity, relative displacement and relative acceleration of A w.r.t B separately in x and y axis. Let $t$ be the time of flight and taking upward direction positive
$\begin{array}{cc}\text{X}& \text{Y}\\ u_{ABx}=8+u_0\cos {45}^{\circ }& u_{ABy}=0-(-u_0\sin {45}^{\circ })\\ & u_{ABy}=u_0\sin {45}^{\circ }\\ a_{ABx}=0& a_{ABy}=g-g=0\\ \text{Using second equation of motion}& \\ S_x=u_{ABx}t+\frac{1}{2}{a}_{AB}{t}^2& \\ 8=(8+u_0\cos {45}^{\circ })t& S_{AB}=-(2-6)=4\text{m}\\ 8=8t+u_0\cos {45}^{\circ }t\dots \dots \left(1\right)& S_{AB}=u_{AB}t\\ & \Rightarrow 4=\left(u_0\sin {45}^{\circ }\right)t\dots \dots \left(2\right)\end{array}$
Solving (1) and (2), we get
$t=0.5\text{s}$ and $u_0=8\sqrt{2}\text{m/s}$
Using first equation of motion, velocity of body $A$ and $B$ after time $t$ can be written as,
$v_A=8\hat{i}+(-gt)\hat{j}$ $v_B=(-8\sqrt{2}\cos {45}^{\circ })\hat{i}+(-8\sqrt{2}\sin {45}^{\circ }-gt)\hat{j}$
On solving, we get
$v_A=9.4\text{m/s};v_B=15.2\text{m/s}$