Question

The point from where a ball is projected is taken as the origin of the coordinates axes. The $x$ and $y$ components of its displacement are given by $x=6t$ and $y=8t-5t^2$. What is the velocity of projection?

• A. $6m{s}^{-1}$
• B. $8m{s}^{-1}$
• C. $10m{s}^{-1}$
• D. $14m{s}^{-1}$

Answer 1

Answer: (C)

$10m{s}^{-1}$

The velocities are given as,

$\frac{dx}{dt}=6$

$v_x=6$

$\frac{dy}{dt}=6$

$v_y=8-10t$

The velocity at projection is given as,

$v\left(t\right)=v_x\hat{i}+v_y\hat{j}$

$v\left(t\right)=6\hat{i}+\left(8-10t\right)\hat{j}$

$v\left(0\right)=6\hat{i}+8\hat{j}$

$\left|v\left(0\right)\right|=\sqrt{6^2+8^2}$

$\left|v\left(0\right)\right|=10m/s$

The velocity of projection is $10m/s$.