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Expert-verified Found in: Page 123 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # Verify the ranges for the projectiles in Figure 3.41 (a) for $\mathbit{\theta }\mathbf{=}\mathbf{45}\mathbf{°}$and the given initial velocities. When the initial velocity is $30\text{m/s}$, the range of the projectile is $91.8\text{m}$.

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

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

See the step by step solution

## 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,

$R=\frac{{u}^{2}\mathrm{sin}\left(2\theta \right)}{g}$ ....(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,$\theta =45.0°$.
• The initial velocity of projection $30\text{m/s}$.
• The range of projectile for $30\text{m/s}$is $91.8\text{m}$.
• The initial velocity of projection $40\text{m/s}$.
• The range of projectile for $40\text{m/s}$is $163\text{m}$.
• The initial velocity of the projectile $50\text{m/s}$.
• The range of projectile for $50\text{m/s}$ is $255\text{m/s}$.

## Step 3: Case (i) Horizontal range

When the initial velocity is $30\text{m/s}$, 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,

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

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

## Step 4: Case (ii) Horizontal range

When the initial velocity is $40\text{m/s}$, 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,

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

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

## Step 5: Case (iii) Horizontal range

When the initial velocity is $50\text{m}/\text{s}$, 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,

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

Hence, when the initial velocity is , the range of the projectile is . ### Want to see more solutions like these? 