Question

A projectile is launched with initial velocity vo. If a second projectile is launched with half of the initial velocity of the first projectile. Compare the range of the second projectile with the first one:

Answer 1

Comparison of the range:

If a second projectile is launched with half of the initial velocity of the first projectile. The range of the second projectile is one-fourth of the first one.

Calculation:

Step-1:

It is given that the first particle has an initial velocity of $v_0$, and the second particle has an initial velocity of $\frac{v_0}{2}$. It is required to compare the range of the second projectile with the first one.

It is known that the range of a projectile is calculated as,

$R=\frac{u^{2} \sin 2 \theta}{g}$

Where u is the initial speed, g is the acceleration due to gravity, and $\theta$ is the angle at which the particle is thrown.

This problem $\theta$ is the same in both cases.

Step-2:

Therefore the range of the first projectile is,

$R_1=\frac{v_0^{2} \sin 2 \theta}{g}$

The range of the second projectile is,

$\begin{aligned}R_2&=\frac{(\frac{v_0}{2})^{2} \sin 2 \theta}{g}\\&=\frac{v_0^2\sin 2 \theta}{4g}\\\end{aligned}$

Therefore,

$\begin{aligned}\\\frac{R_2}{R_1}&=\frac{1}{4}\\\Rightarrow R_2&=\frac{1}{4}R_1\\\end{aligned}$

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