Question

A projectile is fired horizontally with a velocity of $98\text{m/s}$ from the top of a hill $490\text{m}$ high. Find the angle that the line joining the top of the hill and the point on the ground where the projectile hits make with the vertical.
(Take $g=9.8{\text{ms}}^{-2}$)

• A. ${\tan }^{-1}\left(\frac{1}{2}\right)$
• B. ${\tan }^{-1}\left(2\right)$
• C. ${\tan }^{-1}\left(\frac{1}{3}\right)$
• D. ${\tan }^{-1}\left(3\right)$

Answer 1

The correct option is B ${\tan }^{-1}\left(2\right)$

Here we cannot use the formula of range directly.
We will have to deal with the motion along x-axis and y-axis separately.
Let ${}^{\prime }{O}^{\prime }$ be the origin
At $A$, $s_y=490\text{m}$
So, applying second equation of motion,
$s_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$
$\therefore 490=0+\frac{1}{2}\left(9.8\right)t^2$
$\therefore t=10\text{s}$
$BA=s_x=u_{x}t+\frac{1}{2}{a}_x{t}^2$
or $BA=\left(98\right)\left(10\right)+0$
or $BA=980\text{m}$

In $\Delta OBA$
$\tan \theta =\frac{OB}{BA}=\frac{490}{980}$
$\Rightarrow \tan \theta =\frac{1}{2}$
$\Rightarrow \tan \alpha =\tan (90-\theta )$
$\Rightarrow \tan \alpha =2$
$\therefore \alpha ={\tan }^{-1}\left(2\right)$