Question

A body projected horizontally with a speed v not from some height. find the velocity of the body when it covers equal distance horizontal and vertical directions

Answer 1

you mean, A body is projected horizontally with a speed $v_0$ from some height.

for horizontal direction,

initial velocity of body, $u_x=v_0$

acceleration, $a_x=0$

using formula, $x=u_xt+\frac{1}{2}a_xt^2$

or, $x=u_xt$......(1)

velocity of body after time, $v_x=u_x+a_xt$

or, $v_x=u_x+0=v_0$......(2)

for vertical direction

initial velocity of body, $u_y=0$

acceleration , $a_y=-g$

then, $y=u_y+\frac{1}{2}a_yt^2$

or, $y=0-\frac{1}{2}gt^2=\frac{1}{2}gt^2$.....(3)

velocity after time, t $v_y=u_y+a_yt$

=-gt ......(4)

a/c to question, |x| = |y|

or, $v_0t=\frac{1}{2}gt^2$

or, $t=\frac{2v_0}{g}$.....(5)

now velocity of body after time t, can be written as $\vec{v}=v_x\hat{i}+v_y\hat{j}$

=$v_0\hat{i}-2v_0\hat{j}$ [ from equations (2), (4) and (5) ]

hence magnitude of velocity will be $\bf{\sqrt{5}v_0}$