Question

A ball is projected from a certain point on the ground at an angle with the horizontal surface. The horizontal and vertical displacement x and y(in metres) vary with time (in seconds) as
$x=10\sqrt{3}t$
$y=10t-5t^2$
What is the magnitude and direction of the velocity of projection?

• A. $20m{s}^{-1}$ at an angle of ${30}^{\circ }$ with the horizontal
• B. $20m{s}^{-1}$ at angle of ${60}^{\circ }$ with the horizontal
• C. $10m{s}^{-1}$ at angle of ${30}^{\circ }$ with the horizontal
  
• D. $10m{s}^{-1}$ at angle of ${60}^{\circ }$ with the horizontal
  

Answer 1

The correct option is A

$20m{s}^{-1}$ at an angle of ${30}^{\circ }$ with the horizontal

We know that the position coordinates x and y of a projectile are given by
$x=\left(v_{0}cos\theta \right)t........\left(1\right)andy=\left(v_{0}sin\theta \right)t-\frac{1}{2}g{t}^2........\left(2\right)$
Comparing the given equations,$v_{0}cos\theta =10\sqrt{3}m{s}^{-1}$ and $v_{0}sin\theta =10m{s}^{-1}$
Squaring and adding we get, $v_0^2={10}^2+{10}^2\times 3=400$
$\Rightarrow v_0=20m{s}^{-1}$
Also, $tan\theta =\frac{1}{\sqrt{3}}$
$\Rightarrow \theta ={30}^{\circ }.$
Hence, the correct choice is (a).