Question

The two graphs of a given projectile show that variation of vertical velocity component with time and with horizontal displacement. What is the x-compound of velocity in m/s?

Answer 1

$V_Y=u_Y-gt$

At, $V_Y=0$

$u_Y=gt$

$t=\frac{1}{2}s$

$\frac{4y}{g}=\frac{1}{2}$

$u_Y=\frac{g}{2}m/s$

At $x=2m,V_Y=20$ i.e maximum height is achieved

$x=u_{X}t$,at maximum height $V_Y=0,t=\frac{1}{2},x=2m$

$2=u_X.\frac{1}{2}$

$u_X=4m/s$

$V_X=u_X=4m/s$

$V_X=4m/s$