Question

A projectile is shot at an angle of $\dfrac{\pi}{6}$ and a velocity of 48$\dfrac{m}{s}$. How far away will the projectile land?

Answer 1

Hint: The formula for the range of the projectile is used in determining the distance the projectile will cover in its motion. In its motion, the x component of the velocity does not change in the motion, whereas, the y component suffers a deceleration of g first in the ascent then an same acceleration in the descent.
Formula Used:
The Horizontal range of the projectile is given by using the formula:
$R = \dfrac{u^2 \sin 2 \theta}{g}$,
Where u is the initial velocity and $\theta$ is measured with respect to horizontal (x) direction.

Complete answer:
We are given u = 48 m/s and $\theta = \dfrac{\pi}{6}$. Substituting these values in the formula, we get:
$R = \dfrac{(48)^2 \sin 2 \left( \dfrac{\pi}{6} \right)}{10} = 199.5$ m.
Therefore, when we shoot any projectile with initial velocity of 48 m/s making an angle of $\dfrac{\pi}{6}$ with the horizontal, we find that it will land at a distance of about 200 m from the place where it started.

Additional information:
Notice here that no matter how heavy or light the object is, the projectile will land at the same place. Such results are obtained only when the air resistance is completely neglected. If you drop a small metal ball and a light and big plastic ball from the top of a tower, it will land at different times and from our intuition, we say that the small metal ball comes to the ground first but such case takes into account the air resistance and in our calculations we completely ignore air resistance.

Note:
The angle given here is with respect to the horizontal, one should take that into account so one knows that along horizontal direction we get cosine component of the velocity and along vertical direction, we get sine component. The time of ascent and descent is the same and depends entirely on the sine component of the velocity.