Question

A particle is projected at an angle of ${37}^{\circ }$ with an inclined plane. The inclined plane is at an angle of ${60}^{\circ }$ with the horizontal. Find
I. Time of flight of particle.
II. Distance traveled by particle (AB) along the inclined plane

• A. I. Time of flight $=2.4s$
  II. Distance traveled $=5.7m$
• B. I. Time of flight $=1.2s$
  II. Distance traveled $=5.7m$
• C. I. Time of flight $=4.8s$
  II. Distance traveled $=11.4m$
• D. I. Time of flight $=6.4s$
  II. Distance traveled $=11.4m$

Answer 1

The correct option is A

I. Time of flight $=2.4s$
II. Distance traveled $=5.7m$

Here, $u_y=usin\alpha =usin{37}^{\circ }=\frac{3}{5}\times 10=6m/s$
$a_y=-gcos\theta =-gcos{60}^{\circ }=-10\times \frac{1}{2}=-5m/s^2$
Now, from first equation of motion, $0=u_y+a_{y}t\Rightarrow t=\frac{6}{5}$
Hence, time period, $T=2t=2.4s$