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Q31PE

Expert-verifiedFound in: Page 123

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 $\mathit{\theta}\mathbf{=}\mathbf{45}\mathbf{\xb0}$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 $255\text{m}$.

**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.

- The angle of projection is,$\theta =45.0\xb0$.
- 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 $255\text{m/s}$.

When the initial velocity is $30\text{m/s}$, and the angle of projection is $45\xb0$, the horizontal range can be calculated using equation (1.1).

Substitute the value in equation 1.1, and we get,

$\begin{array}{rcl}R& =& \frac{{\left(30\text{m/s}\right)}^{2}\times \mathrm{sin}\left(2\times 45\xb0\right)}{\left(9.8{\text{m/s}}^{\text{2}}\right)}\\ & =& 91.8\text{m}\\ & & \end{array}$

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

When the initial velocity is $40\text{m/s}$, and the angle of projection is $45\xb0$, the horizontal range can be calculated using equation (1.1).

Substitute the value in equation 1.1, and we get,

$\begin{array}{rcl}R& =& \frac{{\left(40\text{m/s}\right)}^{2}\times \mathrm{sin}\left(2\times 45\xb0\right)}{\left(9.8{\text{m/s}}^{2}\right)}\\ & \approx & \text{163}\text{m}\\ & & \end{array}$

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

When the initial velocity is $50\text{m}/\text{s}$, and the angle of projection is $45\xb0$, the horizontal range can be calculated using equation (1.1).

Substitute the value in equation 1.1, and we get,

$\begin{array}{rcl}R& =& \frac{{\left(50\text{m/s}\right)}^{2}\times \mathrm{sin}\left(2\times 45\xb0\right)}{\left(9.8{\text{m/s}}^{2}\right)}\\ & \approx & \text{255}\text{m}\\ & & \end{array}$

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

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