Question

A projectile leaves ground level at an angle of \(68^{\circ}\) above the horizontal. As it reaches its maximum height, \(H\), it has traveled a horizontal distance, \(d\), in the same amount of time. What is the ratio \(H / d ?\)

Short answer

Answer: The ratio \(\frac{H}{d}\) is approximately 1.20.

Step-by-step solution

Analyze the vertical motion

First, let's consider the vertical motion. To analyze the vertical motion, we should find the initial vertical velocity at which the projectile is launched. We can do this using the launch angle and the initial velocity as follows: Let \(v_0\) be the initial velocity (magnitude) and \(\theta\) be the angle above the horizontal, which is given as \(68^{\circ}\). Then, the initial vertical velocity, \(v_{0y}\), can be found using the sine function: \(v_{0y} = v_0 \sin{(\theta)}\) We also know that the maximum height (H) is reached when the vertical velocity becomes zero.

Find the time taken to reach maximum height

Let \(t_H\) represent the time taken to reach maximum height, and \(g\) be the acceleration due to gravity (assuming it is positive downwards). We can use the following kinematic equation to find \(t_H\): \(v_{0y} - gt_H = 0\) Now, we can substitute the expression for \(v_{0y}\) from Step 1 and solve for \(t_H\): \(t_H = \frac{v_0 \sin{(\theta)}}{g}\)

Analyze the horizontal motion

Now let's consider the horizontal motion. The horizontal velocity, \(v_{0x}\), can be found using the cosine function: \(v_{0x} = v_0 \cos{(\theta)}\) The horizontal distance (d) covered can be found by multiplying the horizontal velocity with the time taken to reach maximum height, which equals to \(t_H\): \(d = v_{0x}\cdot t_H\)

Find the ratio H/d

Now we have expressions for both H and d using the equations from Step 2 and Step 3. To find the ratio \(\frac{H}{d}\), we can divide the expressions for H and d: \(\frac{H}{d} = \frac{\frac{v_0^2 \sin^2{(\theta)}}{2g}}{\frac{v_0^2 \sin{(\theta)}\cos{(\theta)}}{g}}\) Now we can simplify the expression: \(\frac{H}{d} = \frac{\sin^2{(\theta)}}{2\sin{(\theta)}\cos{(\theta)}} = \frac{\sin{(\theta)}}{2\cos{(\theta)}}\) Now, we can plug in the given angle \(\theta = 68^{\circ}\): \(\frac{H}{d} = \frac{\sin{(68^{\circ})}}{2\cos{(68^{\circ})}}\) Finally, we can calculate the numerical value of this ratio: \(\frac{H}{d} = \frac{0.8988}{2\cdot 0.3746} \approx 1.20\) Therefore, the ratio \(\frac{H}{d}\) is approximately 1.20.

Key concepts

Kinematic Equations

Understanding kinematic equations is key to solving problems in projectile motion. These equations help describe the motion of objects in terms of their velocity, acceleration, and displacement. They assume constant acceleration, which is usually gravity when it comes to projectiles.

For vertical motion, the initial vertical velocity can be derived using:

• \( v_{0y} = v_0 \sin{(\theta)} \)

For horizontal motion, where there is no acceleration, the initial velocity is:

• \( v_{0x} = v_0 \cos{(\theta)} \)

By combining these, we understand the trajectory and motion of projectiles. Mastery of these equations allows us to predict various aspects like maximum height or distance traveled.

Maximum Height

The maximum height in projectile motion is reached when the vertical velocity becomes zero. At this point, the projectile has climbed as high as it can against gravity before starting to descend.

Using the kinematic equation for vertical motion:

• \( v_{0y} - gt_H = 0 \)

we derive the time to reach maximum height:

• \( t_H = \frac{v_0 \sin{(\theta)}}{g} \)

The formula for maximum height \( H \) is then:

• \( H = \frac{v_0^2 \sin^2{(\theta)}}{2g} \)

This shows that the height depends on the launch speed and angle.

Horizontal Distance

Horizontal distance, also known as range, depends on the time of flight and horizontal velocity. Since horizontal motion is not influenced by gravity, it moves at a constant velocity.

The horizontal component of velocity is given by:

• \( v_{0x} = v_0 \cos{(\theta)} \)

The horizontal distance \( d \) is calculated as:

• \( d = v_{0x} \cdot t_H \)

The greater the initial speed or the more favorable the angle, the greater the range traveled.

Launch Angle

The launch angle is crucial because it determines how the initial velocity is split into vertical and horizontal components.

Angles closer to 45 degrees generally yield the furthest distance for projectiles on level ground, due to the optimal balance between height and distance.

In this specific problem, we have:

• \( \theta = 68^{\circ} \)

This angle affects calculations such as:

• \( \sin{(68^{\circ})} \approx 0.8988 \)
• \( \cos{(68^{\circ})} \approx 0.3746 \)

Thus, understanding and applying \( \sin \) and \( \cos \) as functions of the launch angle is vital for resolving projectile motion problems.