Question

The vertical height of the projectile at time t is given by $y=4t-t^2$ and the horizontal distance covered given by $x=3t$.What is the angle of projection with horizontal?

• A. $ta{n}^{-1}\frac{3}{5}$
• B. $ta{n}^{-1}\frac{4}{5}$
• C. $ta{n}^{-1}\frac{4}{3}$
• D. $ta{n}^{-1}\frac{3}{4}$

Answer 1

The correct option is C $ta{n}^{-1}\frac{4}{3}$

Given,

$x=3t$

$y=4t-t^2$

Initial velocity of projection along x-axis at $t=0sec$ is,

$v_x=\frac{dx}{dt}=3m/s$

Initial velocity of projection along y-axis at $t=0sec$ is

$v_y=\frac{dy}{dt}=4-2t$

$v_y{|}_{t=0sec}=4m/s$

Velocity of the projection, $\overrightarrow{v}=v_x\hat{i}+v_y\hat{j}$

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

Angle of projection, $tan\theta =\frac{v_y}{v_x}$

$tan\theta =\frac{4}{3}$

$\theta =ta{n}^{-1}\frac{4}{3}$

The correct option is C.