Question

a projectile thrown with velocity v0 at an angle x to the horizontal as a range R .It will strick a verticle wall at a distance R/2 from the point of projectile with a speed of​

Answer 1

$\displaystyle\large\boxed {\sf {v=v_0\cos x}}$

Short method:-

The projectile attains maximum height at half of its horizontal range, i.e., at $\displaystyle\sf {\dfrac {R}{2}}$ distance from the point of projection, so that the vertical velocity component becomes zero at that position.

Hence the velocity with which the projectile hits the vertical wall is,

$\displaystyle\longrightarrow\sf {\underline {\underline {v=v_0\cos x}}}$

This is the velocity attained by the projectile at the maximum height, we know it.

Actual method:-

When the projectile travels $\displaystyle\sf {\dfrac {R}{2}}$ distance towards the vertical wall, the horizontal velocity at that point is given by,

$\displaystyle\longrightarrow\sf {(v_x)^2=(v_{0_x})^2+2a_xs_x}$

$\displaystyle\longrightarrow\sf {(v_x)^2=(v_0)^2\cos^2x+2(0)\cdot\dfrac {R}{2}}$

$\displaystyle\longrightarrow\sf {v_x=v_0\cos x}$

The height (vertical displacement, in other words) attained by the projectile at the point is given by,

$\displaystyle\longrightarrow\sf {s_y=\dfrac {R}{2}\tan x-\dfrac {g}{2(v_0)^2\cos^2x}\left (\dfrac {R}{2}\right)^2}$

$\displaystyle\longrightarrow\sf {s_y=\dfrac {R\sin x}{2\cos x}-\dfrac {gR^2}{8(v_0)^2\cos^2x}}$

$\displaystyle\longrightarrow\sf {s_y=\dfrac {R}{2\cos x}\left [\sin x-\dfrac {gR}{4(v_0)^2\cos x}\right]\quad\quad\dots (1)}$

But we know that,

• $\displaystyle\sf {R=\dfrac {(v_0)^2\cdot2\sin x\cos x}{g}}$

Then (1) becomes,

$\displaystyle\longrightarrow\sf {s_y=\dfrac {(v_0)^2\sin x}{g}\left [\sin x-\dfrac {\sin x}{2}\right]}$

$\displaystyle\longrightarrow\sf {s_y=\dfrac {(v_0)^2\sin^2x}{2g}}$

Hence the vertical velocity at that point is given by,

$\displaystyle\longrightarrow\sf {(v_y)^2=(v_{0_y})^2+2a_ys_y}$

$\displaystyle\longrightarrow\sf {(v_y)^2=(v_0)^2\sin^2x-2g\cdot\dfrac {(v_0)^2\sin^2x}{2g}}$

$\displaystyle\longrightarrow\sf {v_y=0}$

Therefore the net velocity with which the projectile hits the vertical wall is,

$\displaystyle\longrightarrow\sf {v=\sqrt{(v_x)^2+(v_y)^2}}$

$\displaystyle\longrightarrow\sf {\underline {\underline {v=v_0\cos x}}}$