Question

If a particle is projected from point A, normal to the plane calculate

(i) Time of flight?

(ii) AB=?

• A. $\frac{T=2sec}{AB=40\sqrt{3}m}$
• B. $\frac{T=1sec}{AB=20\sqrt{3}m}$
• C. Height of point of projection should be given
• D. $\frac{T=4sec}{AB=40\sqrt{3}m}$

Answer 1

The correct option is D

$\frac{T=4sec}{AB=40\sqrt{3}m}$

Resolving along and perpendicular to plane

(i) To calculate the time of flight we have to analyze the motion in y-axis. So we have to use the formula

$y=v_y.t+\frac{1}{2}{a}_y{t}^2$

Here, $y=0,u_y=10\frac{m}{s}$ (As the particle is projected vertically upwards, so whole velocity is along y-axis)

$u_x=0\frac{m}{s}$ (As the particle is projected vertically along y-axis, so there will be no component along x-axis)

Displacement of the particle in y-direction will be zero.

$\therefore 0=10.t-\frac{5}{2}{t}^2\Rightarrow t=4sec$ which is the time taken to cover AB.

(ii) To calculate the distance along AB(range) we have to analyze the motion along x-direction. So we have to use the formula $x=u_x+\frac{1}{2}{a}_x{t}^2$

Here $u_{x}0,a_x=gsin\theta =10sin{60}^o=10\frac{\sqrt{3}}{2}=5\sqrt{3}\frac{m}{s},t=4s$

Here $a_x$ is positive as the particle is coming down in the inclined downward direction.

So, $x=R=0.t+\frac{1}{2}{a}_y{t}^2=\frac{5\sqrt{3}}{2}.(4{)}^2=40\sqrt{3}m$