Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

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Short Answer

Verify the ranges for the projectiles in Figure 3.41 (a) for $θ = 45 °$ and the given initial velocities.

When the initial velocity is $30 m/s$ , the range of the projectile is $91.8 m$ .

When the initial velocity is $40 m/s$ , the range of the projectile is $163 m$ .

When the initial velocity is $50 m/s$ , the range of the projectile is $255 m$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Horizontal range

The horizontal range is the distance between a projectile's point of projection and its point of impact. By adjusting the projectile's initial velocity and projection angle, one may alter the projectile's horizontal range of motion.

The expression for the horizontal range is,

$R = \frac{u^{2} \sin (2 θ)}{g}$ ....(1.1)

Here is the horizontal range, is the angle of projection, and is the acceleration due to gravity.

Step 2: Given data

The angle of projection is, $θ = 45.0 °$ .
The initial velocity of projection $30 m/s$ .
The range of projectile for $30 m/s$ is $91.8 m$ .
The initial velocity of projection $40 m/s$ .
The range of projectile for $40 m/s$ is $163 m$ .
The initial velocity of the projectile $50 m/s$ .
The range of projectile for $50 m/s$ is $255 m/s$ .

Step 3: Case (i) Horizontal range

When the initial velocity is $30 m/s$ , and the angle of projection is $45 °$ , the horizontal range can be calculated using equation (1.1).

Substitute the value in equation 1.1, and we get,

$\begin{array}{rcl} R & = & \frac{{(30 m/s)}^{2} \times \sin (2 \times 45 °)}{(9.8 {m/s}^{2})} \\ = & 91.8 m \end{array}$

Hence, when the initial velocity is $30 m/s$ , the range of the projectile is $91.8 m$ .

Step 4: Case (ii) Horizontal range

When the initial velocity is $40 m/s$ , and the angle of projection is $45 °$ , the horizontal range can be calculated using equation (1.1).

Substitute the value in equation 1.1, and we get,

$\begin{array}{rcl} R & = & \frac{{(40 m/s)}^{2} \times \sin (2 \times 45 °)}{(9.8 {m/s}^{2})} \\ \approx & 163 m \end{array}$

Hence, when the initial velocity is $40 m/s$ , the range of the projectile is $163 m$ .

Step 5: Case (iii) Horizontal range

When the initial velocity is $50 m / s$ , and the angle of projection is $45 °$ , the horizontal range can be calculated using equation (1.1).

Substitute the value in equation 1.1, and we get,

$\begin{array}{rcl} R & = & \frac{{(50 m/s)}^{2} \times \sin (2 \times 45 °)}{(9.8 {m/s}^{2})} \\ \approx & 255 m \end{array}$

Hence, when the initial velocity is , the range of the projectile is .

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College Physics (Urone)

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Short Answer

Verify the ranges for the projectiles in Figure 3.41 (a) for $θ = 45 °$ and the given initial velocities.

Step by Step Solution

Step 1: Horizontal range

Step 2: Given data

Step 3: Case (i) Horizontal range

Step 4: Case (ii) Horizontal range

Step 5: Case (iii) Horizontal range

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College Physics (Urone)

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Short Answer

Verify the ranges for the projectiles in Figure 3.41 (a) for θ=45°and the given initial velocities.

Step by Step Solution

Step 1: Horizontal range

Step 2: Given data

Step 3: Case (i) Horizontal range

Step 4: Case (ii) Horizontal range

Step 5: Case (iii) Horizontal range

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