Question

A projectile leaves ground level at an angle of \(68.0^{\circ}\) 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 of the maximum height reached by the projectile to the horizontal distance it travels in the same amount of time is approximately 1.922 when the projectile leaves ground level at an angle of 68 degrees above the horizontal.

Step-by-step solution

Identify the given information and variables needed to solve the problem

We are given the angle of projection, \(\theta = 68.0^{\circ}\), and asked to find the ratio \(H/d\). We need to find the horizontal distance \(d\) and the maximum height \(H\) in the same amount of time.

Identify the time to reach the maximum height

At the point of maximum height, the vertical component of the projectile's velocity will be zero. So we can write the following equation to find the time, $$v_{yf} = v_{yi} - gt$$ Here, \(v_{yf}\) is the final vertical velocity, \(v_{yi}\) is the initial vertical velocity, \(g\) is the acceleration due to gravity, and \(t\) is time.

Find the initial horizontal and vertical velocities

The initial velocity of the projectile can be split into its horizontal and vertical components using trigonometry, $$v_{xi} = v_{i} \cos{\theta}$$ $$v_{yi} = v_{i} \sin{\theta}$$

Calculate the time to reach the maximum height

We know that \(v_{yf} = 0\) at the maximum height, so the equation becomes: $$0 = v_{yi} - gt$$ $$t = \frac{v_{yi}}{g} = \frac{v_{i} \sin{\theta}}{g}$$

Calculate the horizontal distance \(d\)

The horizontal distance can be found by using the equation: $$d = v_{xi} t = v_{i} \cos{\theta} \left(\frac{v_{i}\sin{\theta}}{g}\right) = \frac{v_{i}^2 \sin{\theta} \cos{\theta}}{g}$$

Calculate the maximum height \(H\)

The maximum height can be found by using the equation: $$H = \frac{1}{2}gt^2$$ Plugging in the value of \(t\) from step 4, we get: $$H = \frac{1}{2}g\left(\frac{v_{i}\sin{\theta}}{g}\right)^2 = \frac{v_{i}^2 \sin^2{\theta}}{2g}$$

Calculate the ratio \(H/d\)

Finally, we can find the ratio \(H/d\) as follows: $$\frac{H}{d} = \frac{\frac{v_{i}^2 \sin^2{\theta}}{2g}}{\frac{v_{i}^2 \sin{\theta} \cos{\theta}}{g}} = \frac{\sin^2{\theta}}{2\sin{\theta} \cos{\theta}} = \frac{\sin{\theta}}{2\cos{\theta}}$$ Now substitute the given value of \(\theta\): $$\frac{H}{d} = \frac{\sin{68.0^{\circ}}}{2\cos{68.0^{\circ}}} \approx 1.922$$ Thus, the ratio \(H/d \approx 1.922\).

Key concepts

Kinematic Equations

Understanding projectile motion is greatly facilitated by the kinematic equations, which describe the motion of objects under constant acceleration. In the context of projectile motion, these equations allow us to predict the projectile's vertical and horizontal position over time.

For instance, the equation \(s = ut + \frac{1}{2}at^2\) represents the distance \(s\) traveled by an object with an initial velocity \(u\), under constant acceleration \(a\), over time \(t\). When analyzing the highest point in a projectile's path, we examine the vertical component of motion, considering gravity as the acceleration, and solve for when the vertical velocity equals zero.

The kinematic framework not only simplifies the calculation of time, distance, and height for projectiles but also extends to many other physical situations where acceleration is a factor.

Trigonometry in Physics

Trigonometry plays a vital role in breaking down complex motion into understandable components. It is particularly crucial in projectile motion, which involves a two-dimensional trajectory.

To analyze the path of a projectile, it is often necessary to resolve the initial velocity into its horizontal and vertical components. This is achieved using the sine and cosine functions of trigonometry, where the angle of projection (\(\theta\)) is a key factor. For example, the initial horizontal velocity (\(v_{xi}\)) can be expressed as \(v_{i} \cos{\theta}\), and the initial vertical velocity (\(v_{yi}\)) as \(v_{i} \sin{\theta}\).

Application in Projectile Motion

By using these trigonometric functions, we can examine the independent motions along the x-axis (horizontal) and the y-axis (vertical), simplifying an otherwise complicated motion into manageable calculations.

Vertical and Horizontal Velocity Components

The trajectory of a projectile is the result of two distinct motions: horizontal (x-axis) and vertical (y-axis). These components are independent of each other, with horizontal motion being affected only by the initial velocity and air resistance (commonly neglected in basic physics problems), and vertical motion being influenced by gravity.

Upon launching a projectile at an angle, its initial velocity vector is split into a horizontal component (\(v_{xi}\)) given by \(v_{i} \cos{\theta}\) and a vertical component (\(v_{yi}\)) represented by \(v_{i} \sin{\theta}\).

• The horizontal velocity remains constant if air resistance is ignored.
• The vertical velocity changes due to gravity, increasing on the way up until it momentarily becomes zero at the peak, and then increasing in the downward direction.

Their distinct behaviors contribute to the parabolic path of the projectile and are essential for calculating the maximum height, time of flight, and horizontal distance traveled.