Question

How does mass affect the horizontal velocity and distance and drop time of a projectile?

Answer 1

Hint: The projectile motion is a classic case of two-dimensional motion. By deriving the expressions for distance travelled, time travelled and horizontal velocity of a projectile, we can arrive at the correlation between these quantities and mass of the projectile.

Complete step by step answer:
A projectile is defined as an object that is in flight after being thrown from ground, at a very high velocity in air. The projectile follows a two-dimensional motion wherein it travels in air to a certain height and drops to ground later, due to the gravity.
The projectile motion follows a parabolic path as shown below:

The initial velocity of the projectile is $u$. Since the velocity is in two-dimensions, we have to consider, at every point of the motion, the horizontal and vertical components of the velocity along the X and Y directions as, ${u_x}$ and ${u_y}$
Consider the three quantities associated with projectile motion as follows: i) Horizontal velocity.
The horizontal component of the velocity $u$ is given by the formula:
${u_x} = u\cos \theta $
where $\theta $ = angle of launch of the projectile
In this expression, we see that the velocity of the projectile in the horizontal direction does not depend on the mass of the projectile.
ii) Time for the projectile to drop.
The time taken for the flight of the projectile is equal to twice the time taken to reach its maximum height. The formula for the time taken for maximum height is given by –
${T_h} = \dfrac{{u\sin \theta }}{g}$
The time taken for the entire flight of the projectile –
$T = 2{T_h} = \dfrac{{2u\sin \theta }}{g}$
Here, we can see that the time only depends on the initial velocity and angle of projection and not the mass of the projectile.
iii) Distance travelled.
The total distance travelled by the projectile or the range of the projectile is equal to total horizontal distance at which the projectile touches ground again, after projection. It is given by the formula –
$R = \dfrac{{{u^2}\sin 2\theta }}{g}$
Here, we can see that the range only depends on the initial velocity and angle of projection and not the mass of the projectile.
Hence, we see that none of these quantities depend on mass.

Note:
There is another important quantity related to a projectile, known as maximum height. The maximum height reached by a projectile is given by the formula –
$H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$
Here, we can see that even the maximum height achieved by the projectile does not depend on the mass of the projectile.