Question

A jet of water is projected at an angle $\theta$ = ${45}^{\circ }$ with horizontal from the point A which is situated at a distance $x$ = $OA$ = $\frac{1}{2}m$ from a vertical wall. If the speed of projection is $v_0=\sqrt{10}m/s$ , find the point $P$ of striking of the water jet with the vertical wall.

• A. =
• B. =
• C. =
• D. =

Answer 1

The correct option is C

=

$u_x$ = $\frac{\sqrt{10}}{\sqrt{2}}$ = $\sqrt{5}$

$s_x$ = $\frac{1}{2}$

Since the ball is hitting the wall after some time and when it hits its t-coordinate will be $\frac{1}{2}$

Now, $a_x$ = $0$

$\Rightarrow$ $\sqrt{5}$ = $\frac{\frac{1}{2}}{t}$

$t$ = $\frac{1}{2\sqrt{5}}$.

so the ball hits the wall after $\frac{1}{2\sqrt{5}}$ sec. How high or low the ball has gone after $\frac{1}{2\sqrt{5}}$ sec from the point of projection is what we need to find.

We know $u_y$ = $\sqrt{5}$

$a_y$ = $-10$

$t$ = $\frac{1}{2\sqrt{5}}$

$s_y$ = $u_{y}t+\frac{1}{2}{a}_y{t}^2$

= $\sqrt{5}\times \frac{1}{2\sqrt{5}}-\frac{1}{2}10\times \frac{1}{4\times 5}$

$s_y$ = $\frac{1}{2}-\frac{1}{4}$ = $\frac{1}{4}m$

So ball hits the wall $\frac{1}{4}m$ above the x-axis

$\Rightarrow$ The coordinates of the point $p$ = $(\frac{1}{2},\frac{1}{4})$.