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Found in: Page 124

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# Derive ${\mathbit{R}}{\mathbf{=}}\frac{{{\mathbf{V}}_{\mathbf{o}}}^{\mathbf{2}}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{2}{\mathbf{\theta }}_{\mathbf{o}}}{\mathbf{g}}$for the range of a projectile on level ground by finding the time at which becomes zero and substituting this value of into the expression for ${{\mathbit{X}}}_{\mathbf{1}\mathbf{}}{{\mathbit{X}}}_{{\mathbf{0}}}$ , noting that ${\mathbit{R}}{\mathbf{=}}{{\mathbit{X}}}_{{\mathbf{1}}}{\mathbf{}}{{\mathbit{X}}}_{{\mathbf{0}}}$.

The range of the projectile motion is derived as required.

See the step by step solution

## Step 1: Definition of initial velocity

When gravity first exerts force on an item, its initial velocity indicates how fast it travels. The final velocity, on the other hand, is a vector number that measures a moving body's speed and direction after it has reached its maximum acceleration.

## Step 2: Stating given data

The velocity can be said to be Vi. The velocity is making some angle with the x axis.

${V}_{iy}={V}_{i}\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}$

The displacement will be $Y=0$ meters.

The time is not given.

The gravitational acceleration is $-9.8m/{s}^{2}$$-9.8m/{s}^{2}$.

## Step 3: Deriving parabolic trajectory of displacement

The initial velocity and the final velocity and the average velocity will be same.

$\overline{V}=\frac{X}{t}\phantom{\rule{0ex}{0ex}}t=\frac{X}{{V}_{i}\mathrm{cos}\theta }..........1$

Now let’s calculate the displacement.

$∆Y={V}_{iy}+\frac{1}{2}{g}_{y}{t}^{2}\phantom{\rule{0ex}{0ex}}0={V}_{i}\mathrm{sin}\theta ×t+\frac{1}{2}\left(-g\right){\left(t\right)}^{2}\frac{1}{2}\left(g\right){\left(t\right)}^{2}={V}_{i}\mathrm{sin}\theta ×t\phantom{\rule{0ex}{0ex}}\frac{1}{2}\left(g\right)\left(t\right)={V}_{i}\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}t=\frac{2{V}_{i}\mathrm{sin}\theta }{g}$

The time when displacement is zero is in the above equation.

The average velocity will be equal to the x component of initial velocity.

$\overline{V}={V}_{ix}={V}_{i}\mathrm{cos}\theta$

Also

$\overline{V}=\frac{∆X}{t}\phantom{\rule{0ex}{0ex}}∆X=\overline{V}×t$

Putting the values of v and t into the above equation, we have

$\overline{V}=\frac{∆X}{t}\phantom{\rule{0ex}{0ex}}∆X={V}_{i}\mathrm{cos}\theta ×\left(\frac{2{V}_{i}\mathrm{sin}\theta }{g}\right)\phantom{\rule{0ex}{0ex}}∆X=\left(\frac{2{{V}_{i}}^{2}\mathrm{sin}\theta \mathrm{cos}\theta }{g}\right)\phantom{\rule{0ex}{0ex}}∆X=\left(\frac{{{V}_{i}}^{2}\left[2\mathrm{sin}\theta \mathrm{cos}\theta \right]}{g}\right)\phantom{\rule{0ex}{0ex}}∆X=\left(\frac{{{V}_{i}}^{2}\left[\mathrm{sin}2\theta \right]}{g}\right)\phantom{\rule{0ex}{0ex}}∆X=\left(\frac{{{V}_{i}}^{2}\left[\mathrm{sin}2\theta \right]}{g}\right)\phantom{\rule{0ex}{0ex}}R=\left(\frac{{{V}_{i}}^{2}\mathrm{sin}2\theta }{g}\right)$

Hence the range of the projectile motion is derived as above.