Tag Archives: projectiles with different initial velocities

College Physics by Openstax Chapter 3 Problem 31


Verify the ranges for the projectiles in Figure 3.40(a) for θ=45º and the given initial velocities.


Solution:

To verify the given values in the figure, we need to solve for individual ranges for the given initial velocities. To do this, we shall use the formula

\text{R}=\frac{\text{v}_{\text{0}}^2\:\sin 2\theta _{\text{0}}}{\text{g}}

When the initial velocity is 30 m/s, the range is

\text{R}=\frac{\left(30\:\text{m/s}\right)^2\:\sin \left(2\times 45^{\circ} \right)}{9.81\:\text{m/s}^2}=91.74\:\text{m}

When the initial velocity is 40 m/s, the range is

\text{R}=\frac{\left(40\:\text{m/s}\right)^2\:\sin \left(2\times 45^{\circ} \right)}{9.81\:\text{m/s}^2}=163.10\:\text{m}

When the initial velocity is 50 m/s, the range is

\text{R}=\frac{\left(50\:\text{m/s}\right)^2\:\sin \left(2\times 45^{\circ} \right)}{9.81\:\text{m/s}^2}=254.84\:\text{m}

Based on the results, we can say that the ranges are approximately equal. The differences are only due to round-off errors.


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