Question

The hang time of a punt is measured to be \(4.50 \mathrm{~s}\). If the ball was kicked at an angle of \(63.0^{\circ}\) above the horizontal and was caught at the same level from which it was kicked, what was its initial speed?

Short answer

The initial speed was approximately 24.7 m/s.

Step-by-step solution

Understand the problem

We need to find the initial speed of the punt. The given hang time is 4.50 seconds. The angle of projection relative to the horizontal is 63.0 degrees. Since the ball is caught at the same level as it was kicked, we can use the symmetry of projectile motion.

Use the time of flight formula

For a projectile motion where the landing height is the same as the launch height, the formula for time of flight (T) is:\[ T = \frac{2u \sin \theta}{g} \]where \( u \) is the initial speed, \( \theta \) is the angle of launch, and \( g \) is the acceleration due to gravity \( (9.8 \ m/s^2) \).

Solve for initial speed

Rearrange the time of flight formula to solve for \( u \) (initial speed):\[ u = \frac{Tg}{2 \sin \theta} \]Substitute \( T = 4.50 \ s \), \( g = 9.8 \ m/s^2 \), and \( \theta = 63.0^{\circ} \) into the formula.

Calculate sine of the angle

Calculate \( \sin 63.0^{\circ} \). It is approximately 0.891.

Perform the calculation

Now substitute all the values into the rearranged formula:\[ u = \frac{4.50 \times 9.8}{2 \times 0.891} \]This calculates to \( u \approx 24.7 \ m/s \).

Key concepts

Initial Speed Calculation

In the world of projectile motion, finding the initial speed is crucial. It helps determine how far and fast an object will move through space. In our exercise, the problem asks for the initial speed of a punt. The ball starts off with an unknown speed and travels until it is caught at the same height from which it was kicked. The key to unlocking the initial speed lies in using the time of flight. We know the hang time, which is the total time the projectile is in the air, is 4.50 seconds.
To find the initial speed, you can use the formula for the time of flight:

• For symmetric projectile motion: \[ T = \frac{2u \sin \theta}{g} \]
• Here, \( T \) is the time of flight (4.50 seconds in this case), \( u \) is the initial speed, \( \theta \) is the angle of launch (63.0 degrees), and \( g \) is the gravitational acceleration (9.8 m/s²).

Rearrange the formula to solve for \( u \): \[ u = \frac{Tg}{2 \sin \theta} \] By substituting the known values into the formula, you can calculate the initial speed of the ball.

Time of Flight

The time of flight is defined as the total time duration for which a projectile remains in motion. It’s a key factor in projectile motion, giving you insight into how long an object remains airborne. In this scenario, the hang time is 4.50 seconds. This time indicates how long the ball is in the air from the moment it is kicked until it lands back at the same height.
Understanding the time of flight helps in calculating other essential parameters of projectile motion, like the initial speed or the range of the projectile. The formula used here is:

• The time of flight for symmetrical projectile motion is \[ T = \frac{2u \sin \theta}{g} \]

Since you already have the time of flight, apart from helping in finding the initial speed, this duration also gives practical insights into how the kick is performed such as how high and far the ball will travel.

Angle of Projection

The angle of projection is the angle at which the projectile is launched relative to the horizontal. It plays a pivotal role in determining the path or trajectory and various components of the projectile's motion. In our exercise, the ball is kicked at an angle of 63.0 degrees.
The angle influences how the initial speed is split between horizontal and vertical components:

• The horizontal component (\( u \cos \theta \) ) affects how far the projectile will travel horizontally.
• The vertical component (\( u \sin \theta \) ) impacts the maximum height and the time of flight.

A steeper angle (close to 90 degrees) generally results in a higher but shorter trajectory, while a smaller angle will extend the reach but decrease height. In projectile calculations, using trigonometric functions like \( \sin \theta \) and \( \cos \theta \) is crucial to computing the different motion components efficiently.