Question

During the 2004 Olympic Games, a shot putter threw a shot put with a speed of \(13.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(43.0^{\circ}\) above the horizontal. She released the shot put from a height of \(2.00 \mathrm{~m}\) above the ground. a) How far did the shot put travel in the horizontal direction? b) How long was it until the shot put hit the ground?

Short answer

a) Calculate the range. b) Determine the time it takes for the shot put to hit the ground.

Step-by-step solution

Break down the initial velocity into components

The initial velocity of the shot put can be divided into horizontal and vertical components. To find these components, we can use trigonometric functions. The horizontal component of the velocity, \(v_{0x}\), can be found using the cosine function, while the vertical component, \(v_{0y}\), can be found using the sine function: \[v_{0x} = v_0 \cos{\theta}\] \[v_{0y} = v_0 \sin{\theta}\] where \(v_0 = 13.0 ~\text{m/s}\) and \(\theta = 43.0^\circ\). Calculate these components.

Apply the equations of motion in the horizontal direction

In the horizontal direction, there is no acceleration. Thus, the horizontal velocity remains constant. The horizontal distance, or range, can be obtained by multiplying the horizontal velocity by the total time: \[R = v_{0x} \cdot t\] where \(R\) is the horizontal distance traveled, \(v_{0x}\) is the horizontal component of the initial velocity, and \(t\) is the time it takes for the shot put to hit the ground. We will calculate the range once we find the time \(t\).

Apply the equations of motion in the vertical direction

In the vertical direction, the shot put is subject to gravitational acceleration. To find the time until it hits the ground, we can use the following equation of motion: \[y = v_{0y} \cdot t - \frac{1}{2}gt^2\] where \(y\) is the vertical displacement from the initial height of the shot put (which is 2.00 m), \(v_{0y}\) is the vertical component of the initial velocity, \(g\) is the acceleration due to gravity (approximately \(9.81 ~\text{m/s}^2\)), and \(t\) is the time it takes for the shot put to hit the ground. Since the shot put hits the ground, \(y = -2.00 ~\text{m}\). Solve for \(t\).

Calculate the horizontal distance

Now that we have found the time until the shot put hits the ground, we can go back to the horizontal equation of motion in Step 2. Plug in the values for \(v_{0x}\) and \(t\) into the equation and calculate the horizontal distance traveled by the shot put. a) The horizontal distance traveled by the shot put is the range, \(R\). b) The time it takes for the shot put to hit the ground is \(t\).

Key concepts

Initial Velocity Components

Understanding the initial velocity components is crucial when analyzing projectile motion physics. The initial velocity, which is the speed the object is given at the starting point, is often at an angle, meaning it has both horizontal and vertical components. To calculate these, we break down the initial velocity vector into two perpendicular vectors using basic trigonometry.

For a projectile launched at an angle \(\theta\), the horizontal component, \(v_{0x}\), is found through \(v_{0x} = v_0 \cos{\theta}\), and the vertical component, \(v_{0y}\), through \(v_{0y} = v_0 \sin{\theta}\). Understanding these components can help visualize the projectile's trajectory, which starts with the velocity pointing in the release direction and evolves as the projectile moves through the air.

Equations of Motion

To predict the path of a projectile, we rely on the equations of motion. These equations describe how the position of the projectile changes over time, taking into account its initial velocity and any acceleration it experiences. In the case of a projectile, the horizontal motion has a constant velocity, as there is no horizontal acceleration (assuming no air resistance). This constancy allows us to predict where the projectile will be at any given time along the horizontal axis.

On the other hand, vertical motion is affected by gravitational acceleration. The vertical velocity alters with time, reflecting the object's rise and fall. The primary equations that define these two motions allow us to calculate the range and time of flight of the projectile, which are key to solving related physics problems.

Gravitational Acceleration

Gravitational acceleration, \(g\), is the acceleration due to Earth's gravity that affects any object in free fall. It is approximately \(9.81 \text{m/s}^2\) and acts downward towards the center of the Earth. When we examine the vertical motion of a projectile, we apply gravitational acceleration to see how it impacts the object's speed and trajectory over time.

The formula that we use in the vertical direction is \(y = v_{0y} \cdot t - \frac{1}{2}gt^2\), where \(y\) is the change in vertical position from the point of release. This equation takes into account the upward motion against gravity and the subsequent downward motion, which is an acceleration towards the Earth, resulting in a parabolic path for the projectile.

Projectile Range Calculation

Calculating the projectile range is an essential aspect of studying projectile motion physics. The range of a projectile is the total horizontal distance it travels during its flight. To calculate the range, we use the horizontal component of the initial velocity and the time of flight.

The range equation is given by \(R = v_{0x} \cdot t\), where \(R\) represents the range, \(v_{0x}\) is the horizontal component of the initial velocity, and \(t\) is the time of flight. As the horizontal motion does not involve any acceleration (discounting air resistance), the horizontal component of the velocity remains constant throughout the flight, making the calculation direct and straightforward.

Projectile Time of Flight

The time of flight of a projectile is the duration it remains airborne. To find this, we can focus on the vertical motion—since the projectile returns to the same vertical position it was launched from (ignoring the difference in initial release height, if any). The upward and downward journeys are symmetrical in time when the launch and landing heights are equal.

Mathematically, the time can be determined using the equation \(y = v_{0y} \cdot t - \frac{1}{2}gt^2\), which, when rearranged to solve for \(t\), allows us to calculate the total time the projectile is in the air. For cases where the projectile lands at a different height from where it was launched, like in the shot put example, the total flight time considers this difference and applies the correct kinematic equation to solve for \(t\). Accurate calculation of the time of flight is crucial for determining other parameters of projectile motion, like the range.