Question

The range of projectile in meters (round off to closest integer) on the inclined plane which is projected perpendicular to the incline plane with velocity $20m/s$ as shown in figure is $(70+x)$(Take $g=10m/s^2$ and $\sin {37}^o=0.6$)

Answer 1

Let $x$ be the direction along the plane and $y$ be the direction perpendicular to the plane
$u_x=0$ m/s
$u_y=20$ m/sec;
$a_x=-gsin{37}^{\circ }$
$a_y=-gcos{37}^{\circ }$

Now, along $y$ direction, total distance traced is $0m$ at the end
So, $h=0$
$h=u_{y}t+\frac{1}{2}{a}_y{t}^2$

$t=\frac{2u_y}{\left(g\cos {37}^{\circ }\right)}$

This is the time of flight of the projectile.

Now along $x$ direction:
$s=u_{x}t+\frac{1}{2}{a}_x{t}^2$

$s=0+\frac{1}{2}\times g\sin {37}^{\circ }{t}^2$

$s=\frac{10\times \left(3/5\right)\times 4\times {20}^2}{2\times {10}^2\times (4/5{)}^2}$

$s=75m$