Question

From Points $A$ and $B$, at height of $2\text{m}$ and $6\text{m}$, two bodies are thrown simultaneously towards each other; one is thrown horizontally with a velocity of $8\text{m/s}$ and the other downwards at an angle of ${45}^{\circ }$ to the horizontal at an initial velocity such that the bodies collide in flight. The horizontal distance between points $A$ and $B$ is $8\text{m}$. Calculate
a) The initial velocity $u_0$ of the body thrown at an angle of ${45}^{\circ }$
b) The times of flight $t$ of the bodies before collision
c) Velocities $v_A$ and $v_B$ of the two bodies at the instant of collision. The trajectories lie in a single plane.

• A. t=1 s, $u_0=10\sqrt{2}\text{m/s}$
  $v_A=9.4\text{m/s};v_B=15.2\text{m/s}$
• B. t=0.5 s, $u_0=8\sqrt{2}\text{m/s}$
  $v_A=9.4\text{m/s};v_B=15.2\text{m/s}$
• C. t=0.5 s, $u_0=8\sqrt{5}\text{m/s}$
  $v_A=5\text{m/s};v_B=10\text{m/s}$
• D. t=0.5 s, $u_0=8\sqrt{2}\text{m/s}$
  $v_A=15.2\text{m/s};v_B=9.4\text{m/s}$

Answer 1

The correct option is B t=0.5 s, $u_0=8\sqrt{2}\text{m/s}$

$v_A=9.4\text{m/s};v_B=15.2\text{m/s}$

Let us analyze the relative velocity, relative displacement and relative acceleration of A w.r.t B separately in x and y axis.
$\begin{array}{cc}\text{X}& \text{Y}\\ u_{ABx}=8+u_0\cos {45}^{\circ }& u_{ABy}=u-(-u_0\sin {45}^{\circ })\uparrow +ve\\ & =u_0\sin {45}^{\circ }\\ a_{ABx}=0& a_{ABy}=0\\ \text{Using second equation of motion}& \\ S_x=u_{ABx}t+\frac{1}{2}{a}_{AB}{t}^2& \\ 8=(8+u_0\cos {45}^{\circ })t& S_{AB}=4\text{m}\uparrow \\ 8=8t+u_0\cos {45}^{\circ }t\dots \dots \left(1\right)& S_{AB}=u_{AB}t\\ & \Rightarrow 4=\left(u_0\sin {45}^{\circ }\right)t\dots \dots \left(2\right)\end{array}$
Solving (i) and (ii), we get
$t=0.5\text{s}$ and $u_0=8\sqrt{2}\text{m/s}$
$(x,y)=(4,0.775)\text{m}$
$V_A=8\hat{i}+(-gt)\hat{j}$ and $V_B=(-8\sqrt{2}\cos {45}^{\circ })t\hat{i}+(-8\sqrt{2}\sin {45}^{\circ }-gt)\hat{j}$
On solving we get
$v_A=9.4\text{m/s};v_B=15.2\text{m/s}$