Question

A ball is thrown horizontally from a cliff such that it strikes ground after $5$ sec. The line of sight from the point of projection to the point of hitting makes an angle of ${37}^0$ with the horizontal. The initial velocity of the projection is $u$ m/sec. Find value of $u/10$ and round off to the nearest integer. (Take $g=10m/s^2$)

Answer 1

Assume $x$ is the direction along the plane and $y$ is the direction perpendicular to the plane
$u_x=u\cos {37}^{\circ }$
$u_y=u\sin {37}^{\circ }$
$a_x=g\sin {37}^{\circ }$
$a_y=-g\cos {37}^{\circ }$

$h=0$ along the $y-$ direction ;
$h=u\sin {37}^{\circ }t+\frac{1}{2}(-g\cos {37}^{\circ })t^2$
$\Rightarrow u\sin {37}^{\circ }=\frac{1}{2}\left(g\cos {37}^{\circ }\right)t$
$\Rightarrow u=5\cot {37}^{\circ }\times 5=33.17$ m/sec
$u/10=3$ (Integer value)