Question

In the figure shown, find time of flight of the projectile along the inclined plane and range OP.

Answer 1

Given that,

Angle $\theta ={45}^0$

Velocity $u=20\sqrt{2}m/s$

We know that,

The time of flight

$t=\frac{2u\sin \theta }{g\cos \theta }$

$t=\frac{2\times 20\sqrt{2}}{10}\times 1$

$t=5.7s$

Now, the horizontal range is

$R=\frac{u^2\sin \theta }{g\cos \theta }$

$R=\frac{20\sqrt{2}\times 20\sqrt{2}}{10}$

$R=80m$

Hence, this is the required solution