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 $y=(8t-5t^2)$ m and $x=6tm$, where t is in seconds. The velocity with which the projectile is projected at t = 0 is:

• A. $8m{s}^{-1}$
• B. $6m{s}^{-1}$
• C. $10m{s}^{-1}$
• D. Not obtainable from the data

Answer 1

The correct option is C $10m{s}^{-1}$

$\text{Step 1: Velocity in x and y direction}$

We know velocity is defined as rate of change of displacement, therefore

Velocity in x direction, $u_x=\frac{dx}{dt}$ $=\frac{d}{dt}6t$ $=6m/s$

Velocity in y direction, $u_y=\frac{dy}{dt}$ $=\frac{d}{dt}(8t-5t^2)$ $=8-10t$

At time $t=0$

$u_y=8m/s$

$\text{Step 2: Velocity of projectile at t=0s}$

Resultant $u=\sqrt{u_x^2+u_y^2}$ $=\sqrt{6^2+8^2}m/s=10m/s$

Hence option $C$ is correct.