Problem 15 - College Physics

Question

A tennis player hits a ball at ground level, giving it an initial velocity of 24 \(\mathrm{m} / \mathrm{s}\) at \(57^{\circ}\) above the horizontal. (a) What are thehorizontal and vertical components of the ball's initial velocity? (b) How high above the ground does the ball go? (c) How long does it take the ball to reach its maximum height? (d) What are the ball's velocity and acceleration at its highest point? (e) For how long a time is the ball in the air? (f) When this ball lands on the court, how far is it from the place where it was hit?

Step-by-Step Solution

Answer

(a) Horizontal: 13.07 m/s, Vertical: 20.13 m/s. (b) Max height: 20.68 m. (c) Time to max height: 2.05 s. (d) Velocity: 13.07 m/s, 0 m/s; Acceleration: 9.8 m/s². (e) Time in air: 4.10 s. (f) Range: 53.57 m.

Step 1: Identify Initial Velocity Components

The initial velocity of the tennis ball is given as 24 m/s at an angle of 57° above the horizontal. We need to determine the horizontal (\(v_{0x}\)) and vertical (\(v_{0y}\)) components of the initial velocity using trigonometric functions.- Horizontal component: \(v_{0x} = v_0 \cos(\theta) = 24 \cos(57^{\circ})\)- Vertical component: \(v_{0y} = v_0 \sin(\theta) = 24 \sin(57^{\circ})\).

Step 2: Calculate Horizontal Component

Calculate the horizontal component using cosine.\[v_{0x} = 24 \cos(57^{\circ}) = 24 \times 0.5446 \approx 13.07 \, \text{m/s}\]

Step 3: Calculate Vertical Component

Calculate the vertical component using sine.\[v_{0y} = 24 \sin(57^{\circ}) = 24 \times 0.8387 \approx 20.13 \, \text{m/s}\]

Step 4: Maximum Height Calculation

To find the maximum height reached by the ball, use the formula for vertical motion. At the maximum height, the vertical component of the velocity becomes zero.\[v_{0y}^2 = 2g h_{max}\]Substitute values:\[\left(20.13\right)^2 = 2 \times 9.8 \times h_{max}\]Solve for \(h_{max}\):\[h_{max} \approx 20.68 \, \text{m}\]

Step 5: Time to Reach Maximum Height

To find the time taken to reach the maximum height, use the equation for vertical velocity:\[v_{y} = v_{0y} - gt\]At the maximum height, \(v_{y} = 0\), so:\[0 = 20.13 - 9.8t\]Solve for \(t\):\[t = \frac{20.13}{9.8} \approx 2.05 \, \text{s}\]

Step 6: Velocity and Acceleration at Maximum Height

At the maximum height, the vertical velocity is 0 m/s. The horizontal velocity remains unchanged at the maximum height, \(v_{0x} = 13.07 \, \text{m/s}\). The acceleration, due to gravity, is \(9.8\, \text{m/s}^2\) downward.

Step 7: Time of Flight

The total time the ball is in the air is twice the time taken to reach the maximum height since the descent takes the same time as the ascent.\[t_{total} = 2 \times 2.05 \approx 4.10 \, \text{s}\]

Step 8: Range Calculation

To calculate how far the ball lands from its initial position, use the range formula for projectile motion:\[R = v_{0x} \times t_{total}\]Substitute values:\[R = 13.07 \times 4.10 \approx 53.57 \, \text{m}\]

Key Concepts

Velocity Components

Projectile motion can be better understood by breaking down the initial velocity of the projectile into components. These components are typically horizontal and vertical.

• **Horizontal Component**: This is determined using the cosine function. For our tennis ball, if it is hit at a speed of 24 m/s and an angle of 57° above the horizontal, the horizontal velocity component can be calculated as:\[v_{0x} = 24 \cos(57^{\circ}) \approx 13.07 \, \text{m/s}\]
• **Vertical Component**: This is calculated using the sine function. Similarly, the vertical velocity component is:\[v_{0y} = 24 \sin(57^{\circ}) \approx 20.13 \, \text{m/s}\]

Understanding these components is crucial because they allow us to predict various elements of projectile motion such as the maximum height, time of flight, and range, each independent of each other in their respective dimensions.

Maximum Height

The maximum height a projectile reaches is the peak point in its vertical motion. At this point, the vertical component of its velocity is zero because gravity has slowed its ascent.To find the maximum height \(h_{max}\), we use the energy conservation concept or kinematic relation:The equation is:\[v_{0y}^2 = 2gh_{max}\]Given our vertical component \(v_{0y} \approx 20.13 \, \text{m/s}\) and gravitational acceleration \(g = 9.8 \, \text{m/s}^2\), we calculate:\[\left( 20.13 \right)^2 = 2 \times 9.8 \times h_{max}\]Solving gives:\[h_{max} \approx 20.68 \, \text{m}\]This height indicates the highest point the ball will reach in its flight. Consideration of gravity's role here is critical because it always acts downwards, affecting only the vertical component of motion.

Time of Flight

The time of flight of a projectile is the total time the projectile remains in the air.This includes the time to rise to the maximum height and then descend back to the ground. The symmetry of projectile motion means the time up is equal to the time down. Therefore, the total time in flight can be obtained by doubling the time to reach the maximum height.We calculated earlier:\[v_{y} = v_{0y} - gt\]at the maximum height, where \(v_{y} = 0\):\[0 = 20.13 - 9.8t\]Solving for time, \(t = \frac{20.13}{9.8} \approx 2.05 \, \text{s}\).Thus, the total time of flight \(t_{total}\) is:\[t_{total} = 2 \times 2.05 \approx 4.10 \, \text{s}\]This calculation assumes no air resistance, and merely gravity acting as acceleration downward at all times.

Range of Projectile

The range of a projectile is the horizontal distance traveled from its initial launch point to where it lands.The range \(R\) can be calculated by multiplying the horizontal velocity component by the total time of flight:\[R = v_{0x} \times t_{total}\]Plugging in our values from earlier steps, where \(v_{0x} \approx 13.07 \, \text{m/s}\) and \(t_{total} \approx 4.10 \, \text{s}\):\[R = 13.07 \times 4.10 \approx 53.57 \, \text{m}\]This range calculation assumes a perfectly level launch and landing area and illustrates how the initial velocity components and overall flight time directly influence the distance traveled.