Problem 63 - College Physics

Question

A shot-putter launches the shot from a vertical distance of \(2.0 \mathrm{~m}\) off the ground (from just above her ear) at a speed of \(12.0 \mathrm{~m} / \mathrm{s} .\) The initial velocity is at an angle of \(20^{\circ}\) above the horizontal. Assume the ground is flat. (a) Compared to a projectile launched at the same angle and speed at ground level, would the shot be in the air (1) a longer time, (2) a shorter time, or (3) the same amount of time? (b) Justify your answer explicitly; determine the shot's range and velocity just before impact in unit vector (component) notation.

Step-by-Step Solution

Answer

(1) Longer time; Range: 12.07 m, Velocity: 11.28i - 6.39j m/s.

Step 1: Analyze Projectile from Ground Level

To identify whether the shot will be in the air longer, shorter, or the same amount of time compared to a launch from ground level, we first calculate the time of flight for a shot put launched at the same speed and angle from the ground (0 height).The vertical component of the initial velocity is calculated as:\( v_{iy} = v_i \sin(\theta) = 12 \sin(20^{\circ}) \approx 4.10 \text{ m/s} \).Using the equation for the time of flight for a projectile launched and landed at the same height:\[ t = \frac{2 v_{iy}}{g} \approx \frac{2 \times 4.10}{9.8} \approx 0.84 \text{ s}, \]where \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity.

Step 2: Analyze Projectile from 2m Height

Now, calculate the time of flight for the shot put launched from 2m vertically above ground level.The shot put starts at position \( y = 2 \) m, with the initial velocity as:\( v_{iy} = 4.10 \text{ m/s} \) (same as before).Apply the kinematic equation for the motion:\[ y = v_{iy} t + \frac{1}{2} (-g) t^2 + y_0, \]where \( y_0 = 2 \) m. Since it returns to the ground, \( y = 0 \):\[ 0 = 4.10t - 4.9t^2 + 2. \]Solving the quadratic equation for \( t \) gives:\[ t = 1.07 \text{ s} \].Since 1.07 s is longer than 0.84 s, the shot is in the air for a longer time.

Step 3: Calculate Horizontal Range

To find the range, use the horizontal component of velocity:\( v_{ix} = v_i \cos(\theta) = 12 \cos(20^{\circ}) \approx 11.28 \text{ m/s} \).Multiplying this by the total time of flight calculated for the 2m height:\[ R = v_{ix} \times t = 11.28 \times 1.07 \approx 12.07 \text{ m}. \]

Step 4: Determine Velocity Before Impact

The velocity just before impact can be found using components:Horizontal component remains the same:\( v_{x} = 11.28 \text{ m/s} \).For vertical component, using \( v_{fy} = v_{iy} - gt \):\( v_{fy} = 4.10 - 9.8 \times 1.07 \approx -6.39 \text{ m/s} \).Thus, in unit vector notation, the velocity just before impact is:\[ \mathbf{v} = 11.28 \mathbf{i} - 6.39 \mathbf{j} \text{ m/s}. \]

Key Concepts

Kinematic Equations

Understanding projectile motion requires the use of kinematic equations, which describe the motion of objects under the influence of gravity. These equations allow us to calculate parameters like time of flight, initial, and final velocities, and displacement. In projectile motion, motion is typically separated into horizontal and vertical components. The time of flight and velocity aspects are handled through these kinematic equations:

• Vertical motion: - The equation: \( y = v_{iy} t + \frac{1}{2} a t^2 \) describes vertical displacement, where \(y\) is the height, \(v_{iy}\) is the initial vertical velocity, \(a\) is acceleration due to gravity, and \(t\) is time.
• Horizontal motion: - Velocity is constant, so: \( x = v_{ix} t \), where \(x\) is horizontal distance and \(v_{ix}\) is horizontal velocity.

By using these equations, we determine how long a projectile stays in air, how far it travels, and how its velocity changes over time.

Initial Velocity

Initial velocity is a critical component in projectile motion, as it determines the projectile's trajectory. It is composed of two parts: the horizontal and vertical components.

• Horizontal component: - Calculated as \( v_{ix} = v_i \cos(\theta) \), where \( v_i \) is the speed and \( \theta \) is launch angle.
• Vertical component: - Calculated as \( v_{iy} = v_i \sin(\theta) \).

For the shot-put exercise, the initial velocity at a speed of 12 m/s and angle of 20° results in a vertical velocity of approximately 4.10 m/s and horizontal velocity of 11.28 m/s. This breakdown helps analyze the separate influences of these components on the projectile's overall motion.

Time of Flight

Time of flight refers to the total time a projectile remains in the air. To calculate it accurately for an elevated launch, you'll need to consider both the initial height and initial vertical velocity. For a projectile launched from a higher point, like the 2 meters elevation in our exercise, the time of flight will differ from launching at ground level. We'll apply vertical kinematic equations, solving:\[ 0 = y = v_{iy} t + \frac{1}{2} (-g) t^2 + y_0 \]\[ 0 = 4.10t - 4.9t^2 + 2 \]This is a quadratic equation, and solving it gives the projectile a time of flight of approximately 1.07 seconds compared to 0.84 seconds if launched from the ground level only. This means the elevated launch increases the air time.

Horizontal Range

The horizontal range is the distance a projectile covers along the horizontal axis. It is influenced by the initial horizontal velocity and the time the projectile is in the air.

• Formula: \[ R = v_{ix} \times t \]
• Here, \(v_{ix}\) is the horizontal component of initial velocity, and \( t \) is the time of flight.

In the exercise, with \( v_{ix} \) approximately 11.28 m/s and a time of flight of 1.07 seconds, the horizontal range is about 12.07 meters. By understanding this calculation, one can predict where a projectile will land, which is essential for optimizing the shot-putter's technique.