Question

What is the maximum vertical height to which a baseball player can throw a ball if he can throw it a maximum distance of \(60.0 \mathrm{~m}\) ? Assume that the ball is released at a height of \(1.60 \mathrm{~m}\) with the same speed in both cases.

Short answer

The maximum vertical height to which the baseball player can throw the ball is 60.60 meters.

Step-by-step solution

Calculate the initial velocity

First the initial velocity required to throw the ball to a distance of 60m needs to be calculated. This can be done using the equation of range in Projectile motion. The equation for range \(R\) is given by \(R = \frac{v^2 \sin(2\theta)}{g}\), where \(v\) is the initial velocity, \(\theta\) is the angle at which the project is launched and \(g\) is the acceleration due to gravity. In this case, it is given that the maximum horizontal distance (which occurs when the launch angle is 45 degrees) is 60m. So, the equation can be rearranged to solve for \(v\), as follows: \(v = \sqrt{\frac{R \cdot g}{\sin(2 \cdot 45^\circ)}} = \sqrt{\frac{60.0 \mathrm{~m} \cdot 9.8 \mathrm{~m/s^2}}{\sin(90^\circ)}} = 34.49 \mathrm{~m/s}\)

Calculate the maximum height

Next we need to calculate the maximum height the ball can attain. The formula for maximum height \(h\) in a projectile motion is \( h = \frac{v^2 \sin^2(\theta)}{2g}\). When calculating the highest point the ball can reach, it's thrown vertically upwards, which means the angle theta will be 90 degrees. But, it's also mentioned in the problem that the ball is released from a height of 1.60m, so we need to take this into account. Substituting the values, we get: \( h = \frac{(34.49 \mathrm{~m/s})^2 \cdot \sin^2(90^\circ)}{2 \cdot 9.8 \mathrm{~m/s^2}} + 1.60 \mathrm{~m} = 60.60 \mathrm{~m}\)

Key concepts

Initial Velocity

Understanding the Concept of Initial Velocity

Initial velocity is a critical component in projectile motion problems. In the context of a baseball throw, it's the speed with which the ball leaves the player's hand. Determining this initial velocity requires understanding that the maximum range of a projectile occurs at a 45-degree launch angle. This idea comes from the equation for range in projectile motion, where the launch angle and speed determine how far the ball will travel.

The formula for the horizontal range \(R\) is \(R = \frac{v^2 \sin(2\theta)}{g}\), where:\

\
• \(v\) stands for initial velocity,\
• \(\theta\) is the launch angle,\
• \(g\) is the acceleration due to gravity (approximately \(9.8 \mathrm{m/s^2}\) on Earth).

Given that the maximum distance and gravity are known elements, we can rearrange this equation to solve for initial velocity as illustrated in the provided problem's solution.

Maximum Height

Exploring the Maximum Height in Projectile Motion

The maximum height a projectile can reach is dictated by both its initial velocity and the angle of launch. In our example with the baseball throw, once you've calculated the initial velocity, the formula \( h = \frac{v^2 \sin^2(\theta)}{2g}\) helps us find the highest point the ball can achieve. This equation takes into account the gravitational pull working against the ball's upward trajectory.

It's important to remember that when a projectile is thrown straight up (at a 90-degree angle to the ground), \(\sin^2(90^\circ)\) equals 1, simplifying the calculation. However, the initial height from which the ball is thrown adds to the resulting maximum height, giving the projectile an extra boost. This additive height must be considered to get an accurate measure of the total maximum height reached, as we did by including the initial release height of \(1.60 \mathrm{m}\).

Angle of Launch

The Significance of Launch Angle in Projectile Motion

The angle at which a projectile is launched plays a pivotal role in determining its trajectory and the distance it covers. This angle is measured from the horizontal ground and affects both the horizontal range and maximum height. In many problems, including the baseball example, different angles yield different projectile behaviors.

For maximum distance, a \(45^\circ\) angle typically provides the ideal compromise between horizontal and vertical components of the initial velocity, taking full advantage of the kinetic energy imparted to the projectile. However, to reach the maximum possible height, a \(90^\circ\) launch angle (vertical launch) is used, as this directs all of the initial kinetic energy to fighting against gravity, with no energy spent on horizontal movement.