Problem 55 - Precalculus with Calculus Previews

Question

Free Throw Under certain conditions the maximum height \(y\) attained by a basketball released from a height \(h\) at an angle \(\theta\) measured from the horizontal with an initial velocity \(v_{0}\) is given by \(y=h+\left(v_{0}^{2} \sin ^{2} \theta\right) / 2 g\), where \(g\) is the acceleration due to gravity. Compute the maximum height reached by a free throw if \(h=\) \(2.15 \mathrm{~m}, v_{\mathrm{o}}=8 \mathrm{~m} / \mathrm{s}, \theta=64.47^{\circ}\), and \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\)

Step-by-Step Solution

Answer

The maximum height reached is 4.785 meters.

Step 1: Understanding the formula

The maximum height formula is expressed as \(y = h + \frac{v_0^2 \sin^2 \theta}{2g}\). This formula shows that the maximum height \(y\) is the initial height \(h\) plus a component that depends on the initial velocity \(v_0\), the angle \(\theta\), and the acceleration due to gravity \(g\).

Step 2: Substituting the given values

Substitute the given values into the formula: \(h = 2.15\) m, \(v_0 = 8\) m/s, \(\theta = 64.47^\circ\), and \(g = 9.81\) m/s². This makes the equation: \[ y = 2.15 + \frac{(8)^2 \sin^2(64.47^\circ)}{2 \times 9.81} \]

Step 3: Calculating \(\sin^2 \theta\)

First, calculate \(\sin(64.47^\circ)\). Using a calculator, \(\sin(64.47^\circ) \approx 0.8988\). Then, calculate \(\sin^2(64.47^\circ) = 0.8988^2 \approx 0.8078\).

Step 4: Calculating the velocity component

Calculate \(v_0^2 \sin^2 \theta\) as follows: \[ 8^2 \times 0.8078 = 64 \times 0.8078 = 51.6992 \]

Step 5: Dividing by the gravitational component

Now divide the velocity component by \(2g\): \[ \frac{51.6992}{2 \times 9.81} = \frac{51.6992}{19.62} \approx 2.635 \]

Step 6: Adding to the initial height

Finally, add this result to the initial height \(h\): \[ y = 2.15 + 2.635 = 4.785 \] m.

Key Concepts

Trigonometric Functions

When you're dealing with projectile motion, trigonometric functions play a significant role, especially sine, cosine, and tangent. In the context of this problem, the angle \(\theta\), which is \(64.47^{\circ}\), helps us determine how high the basketball will reach.

The important trigonometric function here is the sine function. The formula contains \(\sin^2 \theta\), which means you'll need to calculate the sine of the angle and then square it. This results in a value that determines how much the vertical component of the initial velocity \(v_0\) affects the height.

Here are a few useful insights:

• A larger angle will generally give a higher vertical component, increasing the maximum height.
• The trigonometric function is crucial in transforming the horizontal motion into vertical height.
• Using a scientific calculator, you can easily determine \(\sin \theta\) and \(\sin^2 \theta\) by pressing the sin button and then squaring the result.

These steps are essential for accurately utilizing the motion equations.

Kinematics

In physics, kinematics is the branch that describes the motion of objects without considering the cause of motion or the forces that bring it about. This becomes particularly meaningful when analyzing the motion of a projectile like a basketball in free throw.

In this particular problem, the projectile motion equation:
\[ y = h + \frac{v_0^2 \sin^2 \theta}{2g} \]
perfectly encapsulates how various elements of kinematics are in play. Notice how the equation combines initial height \(h\), initial velocity \(v_0\), and the angle of projection \(\theta\) to calculate maximum height.

Key aspects of kinematics useful for this exercise include:

• Each dimension of motion can be analyzed separately. In this case, focus is on the vertical height.
• Both time and forces like gravity influence motion, although gravity is the only force acting on the projectile here.
• Initial velocity and projection angle are pivotal in determining the trajectory and maximum height of the basketball.

Understanding these principles help to navigate through such problems efficiently.

Gravity

Gravity is the force pulling objects toward the Earth and is crucial for understanding projectile motion. In this case, it's the constant \(g = 9.81\, \text{m/s}^2\) that determines how quickly a basketball falls back to the ground.

In physics problems like these, gravity impacts the vertical component of the motion, directly reducing the height that a projectile can reach. The inclusion of gravity in the formula as \(2g\) reveals its role in balancing the initial velocity effect.

Here are some important points about gravity in projectile motion:

• It's a constant force for projectiles that influences their height and time in the air.
• The larger the value of \(g\) in an environment, the shorter the time and height a projectile would achieve.
• Gravity doesn't affect the horizontal component of motion; it only affects the vertical motion, pulling the object down as it travels.

By understanding gravity's impact, one can better appreciate how motion equations predict the behavior of projectiles effectively.