Question

(A) the time of flight of the projectile over the inclined plane
(B) range OP

Answer 1

Given that,

Angle $\theta ={45}^0$

Velocity $v=20m/s$

(a).

Now, the time of flight of the projectile over the inclined plane

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

Now, according to question

$t=\frac{2\times 20\times \frac{1}{\sqrt{2}}}{9.8\times \frac{1}{\sqrt{2}}}$

$t=4.08s$

(b).

Now, range OP

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

According to question

$R=\frac{20\times 20\times \sqrt{2}\times \sqrt{2}}{9.8}$

$R=81.6m$

Hence, the time is $4.08s$ and range is $81.6m$