Problem 14 - College Physics

Question

football is thrown with an initial upward velocity component of 15.0 \(\mathrm{m} / \mathrm{s}\) and a horizontal velocity component of 18.0 \(\mathrm{m} / \mathrm{s}\) . (a) How much time is required for the football to reach the highest point in its trajectory? (b) How high does it get above its release point? (c) How much time after it is thrown does it take to return to its original height? How does this time compare with what you calculated in part (b)? Is your answer reasonable? How far has the football traveled horizontally from its original position?

Step-by-Step Solution

Answer

(a) 1.53 s, (b) 11.49 m high, (c) 3.06 s, distance traveled is 55.08 m. The time is reasonable due to symmetry.

Step 1: Calculate Time to Reach Highest Point

To find the time required for the football to reach the highest point, we need to focus on the vertical motion. At the highest point, the vertical velocity will be zero. We can use the equation for vertical velocity: \[ v = v_0 - gt \]where \( v \) is the final velocity (0 m/s at the highest point), \( v_0 = 15 \, \text{m/s} \) is the initial vertical velocity, and \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.So, \( 0 = 15 - 9.8t \).Solving for \( t \), we get: \[ t = \frac{15}{9.8} \approx 1.53 \, \text{seconds} \].

Step 2: Calculate Maximum Height

To find the maximum height, we'll use the formula for vertical displacement:\[ h = v_0t - \frac{1}{2}gt^2 \]Substituting \( v_0 = 15 \, \text{m/s} \), \( g = 9.8 \, \text{m/s}^2 \), and \( t = 1.53 \, \text{s} \) we found earlier:\[ h = 15 \times 1.53 - \frac{1}{2} \times 9.8 \times (1.53)^2 \]\[ h \approx 22.95 - 11.46 \approx 11.49 \, \text{meters} \].

Step 3: Calculate Total Time to Return to Original Height

The total time for the trip going up and coming back down to the original height is double the time to reach the highest point. Since it takes \( 1.53 \, \text{s} \) to reach the peak, the total time is:\[ 1.53 \times 2 = 3.06 \, \text{seconds} \].

Step 4: Evaluate Reasonableness of Time

The time taken to return to the original height (3.06 s) is twice the time taken to reach the highest point (1.53 s), which is expected since the paths going up and coming down are symmetrical under constant acceleration due to gravity.

Step 5: Calculate Horizontal Distance Traveled

To find out how far the football has traveled horizontally, use the horizontal velocity and the total time in air:\[ d = v_x \times t \]where \( v_x = 18 \, \text{m/s} \) and the total time \( t = 3.06 \, \text{s} \):\[ d = 18 \times 3.06 \approx 55.08 \, \text{meters} \].

Key Concepts

Kinematics

Kinematics is a branch of physics that deals with the motion of objects without considering the forces that cause motion. In projectile motion, such as a football being thrown, kinematics breaks the movement into two components: vertical and horizontal. These components are independent of each other.
The vertical motion is influenced by gravity, affecting how high and how long the object stays in the air. Horizontal motion, on the other hand, is uniform because there are no external forces acting horizontally (ignoring air resistance), making the horizontal velocity constant throughout the flight.
By understanding kinematics, you can predict the path a thrown football will take and calculate key details like maximum height, time of flight, and horizontal distance covered.

Vertical Velocity

Vertical velocity changes throughout the football's flight because it is constantly acted upon by gravity. Initially, when the football is thrown upwards, it has a starting vertical velocity (in our exercise, it's 15 m/s upwards). As it ascends, gravity slows it down until it reaches zero at the peak of its flight.

• At the highest point, the vertical velocity is zero.
• We use the formula: \( v = v_0 - gt \), where \( v_0 \) is the initial vertical velocity and \( g \) is gravity.

After reaching the peak, the ball accelerates downward with the same gravity, picking up speed until it is caught or hits the ground. This symmetrical pattern means the time taken to ascend is equal to the time taken to descend back to the original height if caught at the initial throw height.

Horizontal Displacement

Horizontal displacement is the distance the football travels along the horizontal axis. This motion is independent of the vertical motion and is uniform throughout the flight because there is no horizontal acceleration, assuming air resistance is negligible.
The displacement can be calculated using the formula:
\[ d = v_x \times t \]
where \( v_x \) is the constant horizontal velocity and \( t \) is the total time the football is in the air.
In our example, the velocity is 18 m/s, and the total time, including ascent and descent, is 3.06 seconds, resulting in a horizontal displacement of approximately 55.08 meters.

Acceleration due to Gravity

The acceleration due to gravity is a consistent force affecting all objects in free fall near the Earth's surface. It pulls them towards the ground at approximately 9.8 m/s². This force influences only the vertical component of a projectile's motion, not horizontal.

• When the football is at its highest point, this acceleration causes it to stop ascending.
• Then, it accelerates back downwards, increasing in speed as it returns.

Understanding this concept is crucial when predicting how long a projectile will stay in the air and how it moves vertically. The symmetry in ascent and descent times in projectile motion results directly from this constant acceleration.