Question

If height y and horizontal distance x for a projectile on certain planet (with negligible drag) follow $x=4t$ metre, $y=3t-4t^2$ metre. The initial velocity of projection will be (where t is in second)

• A. $4m{s}^{-1}$
• B. $5m{s}^{-1}$
• C. $6m{s}^{-1}$
• D. $8m{s}^{-1}$

Answer 1

The correct option is B $5m{s}^{-1}$

Given,

$x=4t$

$y=3t-4t^2$

The velocity along the x-axis,

$v_x=\frac{dx}{dt}=4m/s$

The initial velocity at time $t=0sec$ along y-axis,

$v_y=\frac{dy}{dt}$

$v_y=3-8t$

At $t=0sec$

$v_y{|}_{t=0sec}=3m/s$

The initial velocity of projection will be,

$\overrightarrow{v}=v_x\hat{i}+v_y\hat{j}$

$\overrightarrow{v}=4\hat{i}+3\hat{j}$

$|\overrightarrow{v}|=v=\sqrt{(4{)}^2+(3{)}^2}=\sqrt{25}$

$v=5m/s$

The correct option is B.