Question

A jet of water is projected at an angle $\theta ={45}^o$ with horizontal from point A which is situated at a distance $x=OA$ = (a) $1/2$ m, (b) $2$ m from a vertical wall. If the speed of projection is $v_0=\sqrt{1}0m{s}^{-1}$, find point $P$ of striking of the water jet with the vertical wall.

Answer 1

Given,

Velocity of projectile $u=\sqrt{10}m/s$

Angle of projection $\theta ={45}^o$

Range of projectile $R=\frac{u^2\sin 2\theta }{g}=\frac{{\left(\sqrt{1}0\right)}^2\times \sin (2\times {45}^o)}{10}=1m$

Equation of trajectory

$y=x\tan \theta (1-\frac{x}{R})=xtan{45}^o(1-x)=x(1-x)$

$y=x(1-x)$

At $x=\frac{1}{2}=0.5,y=x(1-x)=0.5(1-0.5)=0.25m$

At $x=2,y=x(1-x)=2(1-2)=-1m$

(a) Point $P(x,y)=(0.5,0.25)$ in meters

(b) Point $P(x,y)=(2,-1)$ in meters