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Chapter 3: Problem 70

Show that, for a given initial speed, the horizontal range of a projectile is the same for launch angles \(45^{\circ}+\alpha\) and \(45^{\circ}-\alpha\)

Short Answer

Expert verified
Therefore it is shown that a projectile with a fixed initial velocity will have the same horizontal range for two defined angles: \(45^{\circ} + \alpha\) and \(45^{\circ} - \alpha\).

Step by step solution

01

Understanding the Problem

A projectile motion can be broken down into two independent one-dimensional motions. Along the vertical direction, as the force of gravity influences it, the total time of flight is determined. In the case of horizontal motion, as it is not being acted on by any force (neglecting air resistance), it does not speed up or slow down. The distance covered horizontally depends on the time of flight and the initial speed in the horizontal direction.
02

Understanding the formula

The formula for the range of a projectile on a level plane is given by \( R = \frac{v^2sin2\theta}{g} \), where \(v\) is the initial speed, \(\theta\) is the launch angle and \(g\) is the acceleration due to gravity.
03

Plugging in the angles

We plug \(45^{\circ}+\alpha\) and \(45^{\circ}-\alpha\) into the formula and see that by the properties of sin, \(sin(90^{\circ}+2\alpha) = sin(90^{\circ}-2\alpha)\).
04

Comparing the two formula

Both formulas derived in step 3 are equal, which confirms that for the same initial speed, the range of a projectile launched at \(45^{\circ}+\alpha\) and \(45^{\circ}-\alpha\) will be equal.

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Most popular questions from this chapter

Two projectiles are launched simultaneously from the same point, with different launch speeds and angles. Show that no combination of speeds and angles will permit them to land simultaneously and at the same point.

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