Question

A particle is projected from a point (0,1) on Y-axis (assume + Y direction vertically upwards) aiming towards a point (4,9). It fell on ground along x axis in 1 sec. Taking g = 10 $m/s^2$ and all coordinates in metres. Find the X-coordinate where it fell:

• A. (3,0)
• B. (4,0)
• C. (2,0)
• D. $(2\sqrt{5},0)$

Answer 1

The correct option is C (2,0)


$\tan \theta =\frac{9-1}{4-0}=2$, where $\theta$ is the angle of projection
Displacement in y-direction $S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$
now, $-1=u\sin \theta \left(1\right)-\frac{1}{2}g(1{)}^2$
usin$\theta$ = 4 and from triangle sin $\theta$ =$\frac{2}{\sqrt{5}}\Rightarrow u=2\sqrt{5}m/s$
Displacement in x-direction $S_x=U_{x}t+\frac{1}{2}{a}_x{t}^2$
now $x=ucos\theta \left(1\right)=\left(2\sqrt{5}\right)\times \frac{1}{\sqrt{5}}=2m$