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 \(43^{\circ}\) above the horizontal. She released the shot put from a height of \(2 \mathrm{~m}\) 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

Question: Calculate the horizontal distance traveled by the shot put and the time it took for the shot put to hit the ground, given an initial speed of 13.0 m/s, angle of release of 43 degrees, and initial height of 2 meters. Answer: The shot put traveled approximately 10.31 meters in the horizontal direction, and it took 1.10 seconds for the shot put to hit the ground.

Step-by-step solution

Find the horizontal and vertical components of the velocity

We can find the horizontal and vertical components of the initial velocity using trigonometric functions. The horizontal component (v_x) is found by multiplying the magnitude of the initial velocity by the cosine of the angle of release, and the vertical component (v_y) is found by multiplying the magnitude of the initial velocity by the sine of the angle of release. \(v_x = v\cos(\theta)\) \(v_y = v\sin(\theta)\) Plug the given values: \(v_x = 13.0 \text{ m/s} \cos(43^{\circ})\) \(v_y = 13.0 \text{ m/s} \sin(43^{\circ})\) \(v_x \approx 9.37 \text{ m/s}\) \(v_y \approx 8.86 \text{ m/s}\)

Calculate the time it takes for the shot put to reach the ground

To find the time it takes for the shot put to hit the ground, we can use the vertical motion of the shot put. We can use the following kinematic equation: \(y = y_0 + v_{y0}t - \frac{1}{2}gt^2\) Where \(y\) is the final height (0 m), \(y_0\) is the initial height (2 m), \(v_{y0}\) is the initial vertical velocity, \(g\) is the acceleration due to gravity (9.81 m/s²), and \(t\) is the time. We can solve for \(t\) in this equation. \(0 = 2 + 8.86t - \frac{1}{2}(9.81)t^2\) We can use the quadratic formula to find the value of \(t\): \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) In our case, \(a = -\frac{1}{2}(9.81)\), \(b = 8.86\), and \(c = 2\). Plugging in the values: \(t = \frac{-8.86 \pm \sqrt{(8.86)^2 - 4(-\frac{1}{2}(9.81))(2)}}{2(-\frac{1}{2}(9.81))}\) We get two possible values for \(t\). We discard the negative value since time cannot be negative. So, the final value of \(t\) is: \(t \approx 1.10 \text{ s}\)

Calculate the horizontal distance traveled by the shot put

Now that we have the time it takes for the shot put to hit the ground, we can use the horizontal component of the velocity to find the horizontal distance traveled. Since there is no horizontal acceleration, we can use the simple equation: \(x = x_0 + v_{x0}t\) Where \(x\) is the horizontal distance, \(x_0\) is the initial position (0 m), and \(v_{x0}\) is the initial horizontal velocity. Plug in the values: \(x = 0 + 9.37 \text{ m/s}(1.10 \text{ s})\) \(x \approx 10.31 \text{ m}\) So, the shot put traveled approximately 10.31 meters in the horizontal direction (part a) and it took 1.10 seconds for the shot put to hit the ground (part b).

Key concepts

Kinematic Equations

Kinematic equations are the mathematical formulas used to describe the motion of objects in terms of displacement, velocity, acceleration, and time without considering the forces that cause such motion. These equations are essential when analyzing projectile motion, which is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity.

In the exercise, the kinematic equation for vertical motion, \(y = y_0 + v_{y0}t - \frac{1}{2}gt^2\), was used to calculate the time of flight. Here, \(g\) represents the acceleration due to gravity, which is approximately \(9.81 \text{ m/s}^2\) on Earth. The kinematic equations allow us to separate the motion into vertical and horizontal components, which can be analyzed independently since they are perpendicular to each other.

Trajectory

The term 'trajectory' refers to the path that a projectile follows through space as a function of time. In the context of projectile motion, projectiles follow a curved path known as a parabola due to the influence of gravity on their vertical motion, while the horizontal motion remains constant assuming no air resistance.

The shot putter in the exercise launches the shot put at an angle, creating a two-dimensional motion. The combination of the initial vertical and horizontal velocity components, influenced by gravity, results in a trajectory that peaks and then falls back to the ground. The shape of this trajectory and the maximum height reached are determined by factors such as the initial speed, the angle of release, and the height from which it is released.

Horizontal Distance

Horizontal distance in projectile motion is the total length traveled by the object in the horizontal direction. Since there's no horizontal acceleration (assuming air resistance is negligible), the horizontal distance traveled (also known as horizontal range) can be straightforwardly calculated by multiplying the horizontal component of the initial velocity with the time of flight.

In our exercise, the horizontal component of the velocity (\(v_x\)) is constant because gravity does not influence the horizontal motion. By multiplying this constant velocity by the time it takes for the shot put to land (\(t\)), we find the horizontal range of the shot put. The absence of factors such as wind resistance simplifies this calculation, making the horizontal motion a uniform linear motion.

Time of Flight

The 'time of flight' is the duration that the projectile spends in the air from the moment it is released until the moment it hits the ground. It is an important factor in determining both the horizontal range and the trajectory of the projectile.

To calculate the time of flight for the shot put, the kinematic equation for vertical motion was applied. Since the vertical motion is influenced by gravity, it's considered an accelerated motion, resulting in the quadratic equation that was used. By solving this equation, we find the time during which the shot is airborne. This time factor is crucial for calculating both the peak height reached and the horizontal distance traveled by the projectile.