Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

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Short Answer

Derive $R = \frac{{V_{o}}^{2} s i n 2 θ_{o}}{g}$ 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 $X_{1} X_{0}$ , noting that $R = X_{1} X_{0}$ .

The range of the projectile motion is derived as required.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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 V_i. The velocity is making some angle with the x axis.

$V_{i y} = V_{i} \sin θ$

The displacement will be $Y = 0$ meters.

The time is not given.

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

Step 3: Deriving parabolic trajectory of displacement

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

$\bar{V} = \frac{X}{t} t = \frac{X}{V_{i} \cos θ} . . . . . . . . . . 1$

Now let’s calculate the displacement.

$∆ Y = V_{i y} + \frac{1}{2} g_{y} t^{2} 0 = V_{i} \sin θ \times t + \frac{1}{2} (- g) {(t)}^{2} \frac{1}{2} (g) {(t)}^{2} = V_{i} \sin θ \times t \frac{1}{2} (g) (t) = V_{i} \sin θ t = \frac{2 V_{i} \sin θ}{g}$

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

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

$\bar{V} = V_{i x} = V_{i} \cos θ$

Also

$\bar{V} = \frac{∆ X}{t} ∆ X = \bar{V} \times t$

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

$\bar{V} = \frac{∆ X}{t} ∆ X = V_{i} \cos θ \times (\frac{2 V_{i} \sin θ}{g}) ∆ X = (\frac{2 {V_{i}}^{2} \sin θ \cos θ}{g}) ∆ X = (\frac{{V_{i}}^{2} [2 \sin θ \cos θ]}{g}) ∆ X = (\frac{{V_{i}}^{2} [\sin 2 θ]}{g}) ∆ X = (\frac{{V_{i}}^{2} [\sin 2 θ]}{g}) R = (\frac{{V_{i}}^{2} \sin 2 θ}{g})$

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

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College Physics (Urone)

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Short Answer

Derive $R = \frac{{V_{o}}^{2} s i n 2 θ_{o}}{g}$ 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 $X_{1} X_{0}$ , noting that $R = X_{1} X_{0}$ .

Step by Step Solution

Step 1: Definition of initial velocity

Step 2: Stating given data

Step 3: Deriving parabolic trajectory of displacement

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College Physics (Urone)

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Short Answer

Derive R=Vo2sin2θogfor 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 X1 X0 , noting that R=X1 X0.

Step by Step Solution

Step 1: Definition of initial velocity

Step 2: Stating given data

Step 3: Deriving parabolic trajectory of displacement

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Derive $R = \frac{{V_{o}}^{2} s i n 2 θ_{o}}{g}$ 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 $X_{1} X_{0}$ , noting that $R = X_{1} X_{0}$ .