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Chapter 3: Problem 64

Show that for a projectile launched at an angle of \(45^{\circ}\) the maximum height of the projectile is one quarter of the range (the distance traveled on flat ground).

Short Answer

Expert verified
Answer: When a projectile is launched at a 45-degree angle, its maximum height is one quarter of its range.

Step by step solution

01

Define the initial conditions

Given a launch angle of \(45^{\circ}\), we can write the initial velocity components as follows: - Initial horizontal velocity: \(v_{0x} = v_0 \cos 45^{\circ} = \frac{v_0}{\sqrt{2}}\) - Initial vertical velocity: \(v_{0y} = v_0 \sin 45^{\circ} = \frac{v_0}{\sqrt{2}}\)
02

Obtain expressions for the maximum height and range

To find the expressions for maximum height (\(h_{max}\)) and range (\(R\)), we'll use the equations for displacement with uniform acceleration: - Maximum height: At maximum height, the vertical velocity becomes zero, i.e., \(v_y = 0\). We will use the following equation: \(v_y^2 = v_{0y}^2 - 2gh_{max}\) - Range: To calculate the range, we need to find the time of flight (\(t_f\)) and then use it to calculate the horizontal displacement. The time of flight can be found using the vertical motion equation: \(v_y = v_{0y} - gt_f\)
03

Calculate the maximum height

At maximum height, \(v_y = 0\). So, using the equation from step 2 and replacing \(v_{0y}\) with its value from step 1, we get: \(0 = (\frac{v_0}{\sqrt{2}})^2 - 2g h_{max}\) Solve for \(h_{max}\): \(h_{max} = \frac{v_0^2}{4g}\)
04

Calculate the range

Using the vertical motion equation from step 2 and replacing \(v_{0y}\) and \(v_y\) with their values from step 1, we get: \(0 = \frac{v_0}{\sqrt{2}} - g t_f\) Now, solve for \(t_f\): \(t_f = \frac{v_0}{\sqrt{2}g}\) The horizontal displacement equation is: \(R = v_{0x} t_f\) Using values of \(v_{0x}\) and \(t_f\) from steps 1 and 4: \(R = \frac{v_0}{\sqrt{2}} \cdot \frac{v_0}{\sqrt{2}g} = \frac{v_0^2}{2g}\)
05

Prove that the maximum height is one quarter of the range

Now, let's compare the expressions for \(h_{max}\) and \(R\): \(\frac{h_{max}}{R} = \frac{\frac{v_0^2}{4g}}{\frac{v_0^2}{2g}} = \frac{1}{2}\) Thus, the maximum height is one quarter of the range when the launch angle is \(45^{\circ}\).

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Most popular questions from this chapter

A locust jumps at an angle of \(55.0^{\circ}\) and lands \(0.800 \mathrm{m}\) from where it jumped. (a) What is the maximum height of the locust during its jump? Ignore air resistance. (b) If it jumps with the same initial speed at an angle of \(45.0^{\circ},\) would the maximum height be larger or smaller? (c) What about the range? (d) Calculate the maximum height and range for this angle.
Two angles are complementary when their sum is \(90.0^{\circ}\) Find the ranges for two projectiles launched with identical initial speeds of $36.2 \mathrm{m} / \mathrm{s}$ at angles of elevation above the horizontal that are complementary pairs. (a) For one trial, the angles of elevation are \(36.0^{\circ}\) and \(54.0^{\circ} .\) (b) For the second trial, the angles of elevation are \(23.0^{\circ}\) and \(67.0^{\circ} .\) (c) Finally, the angles of elevation are both set to \(45.0^{\circ} .\) (d) What do you notice about the range values for each complementary pair of angles? At which of these angles was the range greatest?
After being assaulted by flying cannonballs, the knights on the castle walls ( \(12 \mathrm{m}\) above the ground) respond by propelling flaming pitch balls at their assailants. One ball lands on the ground at a distance of 50 m from the castle walls. If it was launched at an angle of \(53^{\circ}\) above the horizontal, what was its initial speed?
The range \(R\) of a projectile is defined as the magnitude of the horizontal displacement of the projectile when it returns to its original altitude. (In other words, the range is the distance between the launch point and the impact point on flat ground.) A projectile is launched at \(t=0\) with initial speed \(v_{i}\) at an angle \(\theta\) above the horizontal. (a) Find the time \(t\) at which the projectile returns to its original altitude. (b) Show that the range is \(R=\frac{v_{\mathrm{i}}^{2} \sin 2 \theta}{g}\) [Hint: Use the trigonometric identity $\sin 2 \theta=2 \sin \theta \cos \theta .]$ (c) What value of \(\theta\) gives the maximum range? What is this maximum range?
A motor scooter rounds a curve on the highway at a constant speed of $20.0 \mathrm{m} / \mathrm{s} .$ The original direction of the scooter was due east; after rounding the curve the scooter is heading \(36^{\circ}\) north of east. The radius of curvature of the road at the location of the curve is $150 \mathrm{m}$ What is the average acceleration of the scooter as it rounds the curve?
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