Question

The height $y$ and the distance $x$ along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by $y=(8t-5t^2)$ metre and $x=6t$ metre where $t$ is in seconds. The velocity of projection is:

• A. $8m/sec$
• B. $6m/sec$
• C. $10m/sec$
• D. $15m/sec$

Answer 1

The correct option is C $10m/sec$

$Y=8t-5t²$

differentiate wrt time

$dy/dt=8-10t$

we know, change in position per unit time is called, velocity .

let $\frac{dy}{dt}=V_y$

$V_y=8-10t$

now at $t=0$ ( initially )

$V_y=8-10\times 0=8$

Also, $x=6t$

differentiate wrt time

$dx/dt=6$

similarly , $dx/dt=V_x=6$

now,

$V=V_x\hat{i}+V_y\hat{j}$

= $8\hat{i}+6\hat{j}$

magnitude of velocity =$|V|=\sqrt{8²+6²}=10$m/s