Question

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

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

Step-by-step solution

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: \(v_{0y} = v_0 \sin 45^{\circ} = \frac{v_0}{\sqrt{2}}\)

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: \(v_y = v_{0y} - gt_f\)

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 \(h_{max}\): \(h_{max} = \frac{v_0^2}{4g}\)

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: \(R = \frac{v_0}{\sqrt{2}} \cdot \frac{v_0}{\sqrt{2}g} = \frac{v_0^2}{2g}\)

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 \(45^{\circ}\).