Question

For a particle projected from an inclined plane as shown in the figure, choose the correct statement(s) regarding the projectile.

• A. Time of flight of projectile is $3\text{s}$
• B. Range of projectile $\left(OP\right)$ is $80\sqrt{3}\text{m}$
• C. Range of projectile $\left(OP\right)$ is $80\sqrt{2}\text{m}$
• D. Time of flight of projectile is $4\text{s}$

Answer 1

Answer: (C), (D)

The correct options are
C Range of projectile $\left(OP\right)$ is $80\sqrt{2}\text{m}$
D Time of flight of projectile is $4\text{s}$

Taking the $X-$axis along the inclined plane and $Y-$axis perpendicular to the inclined plane.


Along $Y-$axis,
$u_y=20\sin 45=\frac{20}{\sqrt{2}}=10\sqrt{2}\text{m/s}$
$a_y=-g\cos 45=\frac{-10}{\sqrt{2}}=-5\sqrt{2}{\text{m/s}}^2$

$s_y=0$ [Particle returns to inclined plane, hence displacement along Y-axis is zero]
$\Rightarrow s_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$
$\Rightarrow 0=10\sqrt{2}t-\frac{1}{2}\times 5\sqrt{2}\times t^2$
$\Rightarrow t=\frac{10\sqrt{2}\times 2}{5\sqrt{2}}=4\text{s}$
$\therefore T=4\text{s}...\left(1\right)$ (Time of flight)

Along $X-$axis,
$u_x=20\cos {45}^{\circ }=20\times \frac{1}{\sqrt{2}}=10\sqrt{2}\text{m/s}$
$a_x=10\sin {45}^{\circ }=10\times \frac{1}{\sqrt{2}}=5\sqrt{2}{\text{m/s}}^2$
Range of the particle, $s_x=OP$
Applying $s_x=u_{x}t+\frac{1}{2}{a}_x{t}^2$
$\Rightarrow OP=10\sqrt{2}\times 4+\frac{1}{2}\times 5\sqrt{2}\times 16$
$\therefore OP=40\sqrt{2}+40\sqrt{2}=80\sqrt{2}\text{m}$
Time of flight of the projectile is $4\text{s}$ and its range is $80\sqrt{2}\text{m}$.