91Ó°ÊÓ

A projectile is launched with speed \(v_{0}\) at an angle \(\alpha_{0}\) above the horizontal. The launch point is a height \(h\) above the ground. (a) Show that if air resistance is ignored, the horizontal distance that the projectile travels before striking the ground is $$ x=\frac{v_{0} \cos \alpha_{0}}{g}\left(v_{0} \sin \alpha_{0}+\sqrt{v_{0}^{2} \sin ^{2} \alpha_{0}+2 g h}\right) $$ Verify that if the launch point is at ground level so that \(h=0\) , this is equal to the horizontal range \(R\) found in Example \(3.8 .\) (b) For the case where \(v_{0}=10 \mathrm{m} / \mathrm{s}\) and \(h=5.0 \mathrm{m},\) graph \(x\) as a function of launch angle \(\alpha_{0}\) for values of \(\alpha_{0}\) from \(0^{\circ}\) to \(90^{\circ} .\) Your graph should show that \(x\) is zero if \(\alpha_{0}=90^{\circ},\) but \(x\) is nonzero if \(\alpha_{0}=0\) ; explain why this is so. (c) We saw in Example 3.8 that for a projectile that lands at the same height from which it is launched, the horizontal range is maximum for \(\alpha_{0}=45^{\circ} .\) For the case graphed in part \((b),\) is the angle for maximum horizontal distance equal to, less than, or greater than \(45^{\circ} 7\) (This is a general result for the situation where a projectile is launched from a point higher than where it lands.)

Step by step solution

01

Understand the problem

We need to calculate the horizontal distance \(x\) traveled by a projectile launched at an angle \(\alpha_0\) and verify a formula for it when \(h=0\). Then, analyze how this distance changes with different launch angles when the launch point is not on the ground.

02

Break down the projectile's motion equations

Separate the motion into horizontal and vertical components. The horizontal motion is characterized by no acceleration, and the initial velocity is \(v_0 \cos \alpha_0\). The vertical motion occurs under gravity, so the vertical velocity initially is \(v_0 \sin \alpha_0\), and the displacement in the vertical direction is influenced by \(-g\).

03

Derive the formula for time of flight

For the projectile to hit the ground from height \(h\), the vertical displacement \(y\) is \(-h\) when \(t\) is the total time of flight. Using the equation for vertical displacement, \(y = v_0 \sin \alpha_0 \cdot t - \frac{1}{2} g t^2\), and setting \(y = -h\), solve for \(t\).

04

Solve for horizontal distance

Using the time of flight \(t\) obtained from the vertical motion step, substituting in the equation \(x = v_0 \cos \alpha_0 \cdot t\) gives the horizontal distance traveled. Simplify to obtain the given formula.

05

Simplification for \(h=0\)

Set \(h=0\) in the derived formula and show that it simplifies to \(R\), the formula for horizontal range where the launch and landing points are at the same height.

06

Plot the horizontal distance graph

For part (b), plot the graph of \(x\) as a function of \(\alpha_0\) from \(0^\circ\) to \(90^\circ\) while keeping \(v_0 = 10\, \text{m/s}\) and \(h = 5\, \text{m}\). Use computational tools to graph the function derived in earlier steps, clearly indicating significant behavior such as when \(x=0\).

07

Analyze maximum range angle

For part (c), refer to the graph generated in the previous step. Determine the angle \(\alpha_0\) for which \(x\) is maximal. Explain why this angle differs from \(45^\circ\), the maximum range angle for a projectile landing at the same height it was launched.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Horizontal Range

The horizontal range of a projectile is the total distance it travels along the horizontal axis before hitting the ground. In projectile motion, air resistance is typically ignored to simplify calculations. This allows us to use a standard formula for the horizontal distance. In the described exercise, it is \[x = \frac{v_{0} \cos \alpha_{0}}{g}\left(v_{0} \sin \alpha_{0}+\sqrt{v_{0}^{2} \sin^{2} \alpha_{0}+2gh}\right)\]Here:

• \(v_{0}\) is the initial velocity.
• \(\alpha_{0}\) is the launch angle.
• \(g = 9.81\, \text{m/s}^2\) is the acceleration due to gravity.
• \(h\) is the initial vertical height from which the projectile is launched.

Remember, if the launch point is at the same level as the landing point, the formula simplifies to the usual horizontal range \(R\) as given for \(h=0\). The range depends on both the speed and angle of the launch. Adjusting these variables helps in understanding how far a projectile can travel horizontally.

Launch Angle

The launch angle \(\alpha_{0}\) is critical in determining how a projectile moves through space. It is measured in degrees from the horizontal and affects both the vertical and horizontal components of the motion.

• When the angle is low, closer to \(0^{\circ}\), the projectile travels more in the horizontal direction but reaches a lower height.
• At \(45^{\circ}\), for launch and landing at the same height, the horizontal range reaches its maximum.
• At \(90^{\circ}\), the projectile goes straight up and comes directly down, covering zero horizontal distance.

In scenarios where the launch point is elevated, as in our exercise with \(h = 5\, \text{m}\), the optimal angle for maximum range is less than \(45^{\circ}\). This is due to the additional height allowing more horizontal travel before descending.

Vertical Displacement

Vertical displacement concerns how high or low the projectile travels relative to its starting point. The starting height \(h\) influences this significantly. Here, the projectile is launched from a height above the ground, which affects how long it will be airborne before reaching the ground.The equation for the vertical displacement is given by:\[y = v_{0} \sin \alpha_{0} \cdot t - \frac{1}{2} g t^{2}\]- \(y\) is set to \(-h\) when solving for the total time the projectile stays in the air.- The higher the launch height, the further the projectile may travel as it takes longer to reach the ground.This is crucial for determining the time of flight, which is next, as it defines how long the projectile remains in the air. The vertical component influences the flight duration, affecting both horizontal range and launch dynamics.

Time of Flight

The time of flight is how long the projectile remains in the air before it lands. Calculating this time is essential, as it directly impacts the horizontal distance traveled. The time is determined by ensuring the total vertical displacement considers the launch height and the effect of gravity.At launch height \(h\), the total time \(t\) is derived from:\[-t = \frac{v_{0} \sin \alpha_{0} - \sqrt{(v_{0} \sin \alpha_{0})^{2} + 2gh}}{g}\]

• This total time helps in computing the horizontal range because the range \(x\) is determined by multiplying this time by the horizontal velocity component, \(v_{0} \cos \alpha_{0}\).
• Longer times in the air generally mean greater horizontal distances.
• At higher angles, it spends more time aloft but less distance horizontally, while at lower angles, it travels faster horizontally with shorter air-time.

Exploring the interplay between time and angle provides a comprehensive view of projectile dynamics.