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Q49PE

Expert-verifiedFound in: Page 124

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Derive ${\mathit{R}}{\mathbf{=}}\frac{{{\mathbf{V}}_{\mathbf{o}}}^{\mathbf{2}}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{2}{\mathbf{\theta}}_{\mathbf{o}}}{\mathbf{g}}$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 ${{\mathit{X}}}_{\mathbf{1}\mathbf{}}{{\mathit{X}}}_{{\mathbf{0}}}$ , noting that ${\mathit{R}}{\mathbf{=}}{{\mathit{X}}}_{{\mathbf{1}}}{\mathbf{}}{{\mathit{X}}}_{{\mathbf{0}}}$. **

The range of the projectile motion is derived as required.

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.

The velocity can be said to be V_{i}. 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}$.

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.

$\u2206Y={V}_{iy}+\frac{1}{2}{g}_{y}{t}^{2}\phantom{\rule{0ex}{0ex}}0={V}_{i}\mathrm{sin}\theta \times t+\frac{1}{2}(-g){\left(t\right)}^{2}\frac{1}{2}\left(g\right){\left(t\right)}^{2}={V}_{i}\mathrm{sin}\theta \times t\phantom{\rule{0ex}{0ex}}\frac{1}{2}\left(g\right)\left(t\right)={V}_{i}\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}t=\frac{2{V}_{i}\mathrm{sin}\theta}{g}$

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

$\overline{V}=\frac{\u2206X}{t}\phantom{\rule{0ex}{0ex}}\u2206X=\overline{V}\times t$

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

$\overline{V}=\frac{\u2206X}{t}\phantom{\rule{0ex}{0ex}}\u2206X={V}_{i}\mathrm{cos}\theta \times \left(\frac{2{V}_{i}\mathrm{sin}\theta}{g}\right)\phantom{\rule{0ex}{0ex}}\u2206X=\left(\frac{2{{V}_{i}}^{2}\mathrm{sin}\theta \mathrm{cos}\theta}{g}\right)\phantom{\rule{0ex}{0ex}}\u2206X=\left(\frac{{{V}_{i}}^{2}\left[2\mathrm{sin}\theta \mathrm{cos}\theta \right]}{g}\right)\phantom{\rule{0ex}{0ex}}\u2206X=\left(\frac{{{V}_{i}}^{2}\left[\mathrm{sin}2\theta \right]}{g}\right)\phantom{\rule{0ex}{0ex}}\u2206X=\left(\frac{{{V}_{i}}^{2}\left[\mathrm{sin}2\theta \right]}{g}\right)\phantom{\rule{0ex}{0ex}}R=\left(\frac{{{V}_{i}}^{2}\mathrm{sin}2\theta}{g}\right)$

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

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