Question

The vertical height of the projectile at time $t$ is given by $y=4t-5t^2$ and the horizontal distance covered is given by $x=3t$ where $x$ and $y$ are in $\text{m}$ and $t$ in $\text{s}$. What is the angle of projection with the horizontal?

• A. ${\tan }^{-1}\left(3/5\right)$
• B. ${\tan }^{-1}\left(4/5\right)$
• C. ${\tan }^{-1}\left(4/3\right)$
• D. ${\tan }^{-1}\left(3/4\right)$

Answer 1

The correct option is C ${\tan }^{-1}\left(4/3\right)$

Displacement in vertical direction at time $t$,
$y=4t-5t^2$
$\Rightarrow u_y=\frac{dy}{dt}=4-10t$
Initial velocity at $t=0$
$u_y=4-0=4\text{m/s}$
Horizontal displacement at time $t$,
$x=3t$
$\Rightarrow v_x=\frac{dx}{dt}=3$
Initial velocity at $t=0$
$u_x=3\text{m/s}$


Now, angle of projection with horizontal,
$\tan \theta =\frac{u_y}{u_x}=\frac{4}{3}$
$\Rightarrow \theta ={\tan }^{-1}\left(4/3\right)$