Question

Find range of projectile on the inclined plane which is projected perpendicular to the incline plane with velocity $20\text{m/s}$ as shown in the figure.

Answer 1

Considering the $\text{x-}$axis parallel to the inclined plane, the line diagrams for acceleration and velocity are as drawn below:

Initial velocity along the $\text{x-}$axis, $u_x=0$

Initial velocity along the $\text{y-}$axis, $u_y=20\text{m/s}$

Now, the acceleration along the $\text{x-}$axis, $a_x=g\sin {37}^{\circ }$

Acceleration along the $\text{y-}$axis,
$a_y=g\cos {37}^{\circ }$

To calculate the time of flight $T$, the displacement along the $\text{y}-$axis has to be taken zero.i.e., $y=0$

So, applying the equation of motion along $\text{y-}$axis,

$y=u_{y}T-\frac{1}{2}{a}_y{T}^2$

Substituting the values,

$\Rightarrow 0=20\times T-\frac{1}{2}\times g\cos {37}^{\circ }\times T^2$

$\Rightarrow 0=20\times T-\frac{1}{2}\times 10\times \frac{4}{5}\times T^2$

$\Rightarrow T=\frac{2\times 20}{10\times \frac{4}{5}}$

$\Rightarrow T=5\text{sec}$

During the same time , distance along $\text{x}-$ axis gives the range of the projectile motion, so the range is calculated by using,

$R=u_{x}T+\frac{1}{2}{a}_x{T}^2$

Substituting the values, we get

$R=0+\frac{1}{2}g\sin {37}^o\times 5^2$

$\Rightarrow R=\frac{1}{2}\times 10\times \frac{3}{5}\times 25$

$\therefore R=75\text{m}$

Accepted answer : $75,75.0,75.00$.