Question

Two particles $A$ and $B$ are projected simultaneously form the two towers of height $10m$ and $20m$ respectively. Particle $A$ is projected with an initial speed of $10\sqrt{2}$ at an angle at ${45}^{\circ }$ with horizontal, while particle $B$ is projected horizontally with speed $10m/s$. If they collide in air, what is the distance d between the towers?

Answer 1

The height of particle A at any instant $t$ is given as,

$h_A=10+u_{A}sin{45}^{0}t-\frac{1}{2}g{t}^2$

$h_A=10+10\sqrt{2}sin{45}^{0}t-\frac{1}{2}g{t}^2=10+10t-\frac{1}{2}g{t}^2$

Similarly, height of particle B at any instant $t$ is given as,

$h_B=20-\frac{1}{2}g{t}^2$

At the collision, both the heights are equal.

$\therefore 10+10t-\frac{1}{2}g{t}^2=20-\frac{1}{2}g{t}^2⟹t=1s$

Horizontal distance moved by particle A in 1 second is,

$x_A=u_A\cos {45}^o\times t=10\sqrt{2}cos{45}^0\times 1=10m$

Horizontal distance moved by particle B in 1 second is,

$x_B=u_B\times t=10\times 1=10m$

$\therefore d=x_A+x_B=10m+10m=20m$