Question

If the pitcher's mound is \(1.25 \mathrm{ft}\) above the baseball field, can a pitcher release a fast ball horizontally at \(92.0 \mathrm{mi} / \mathrm{h}\) and still get it into the strike zone over the plate \(60.5 \mathrm{ft}\) away? Assume that, for a strike, the ball must fall at least \(1.30 \mathrm{ft}\) but no more than \(3.60 \mathrm{ft}\).

Short answer

Yes, the pitcher can deliver the fastball within the strike zone at a speed of 92.0 mph.

Step-by-step solution

Convert Units

Before any calculations, convert all the units to be consistent. Let's convert the initial speed of the ball from miles per hour to feet per second by using the conversion factor: \(1 \mathrm{mi} = 5280 \mathrm{ft}\) and \(1 \mathrm{h} = 3600 \mathrm{s}\). So, \(92.0 \mathrm{mi/h} = 92.0 ×(5280/3600) \mathrm{ft/s} = 134.67 \mathrm{ft/s}\).

Compute Time for the Ball to Fall

Since the ball is projected horizontally, it experiences only gravitational acceleration, which is \(g = 32.2 \mathrm{ft/s^2}\). The time to fall from a height \(h\) is given by \(t = \sqrt{(2h/g)}\). Compute the times to fall \(1.25 \mathrm{ft}\) and \(3.6 \mathrm{ft}\), which gives \(t_1 = \sqrt{(2*1.25/32.2)} \mathrm{s} = 0.35 \mathrm{s}\) and \(t_2 = \sqrt{(2*3.6/32.2) \mathrm{s}} = 0.60 \mathrm{s}\).

Determine the Horizontal Distance Traveled

The fastball travels this distance while it is in the air. Therefore, the horizontal distance is given by \(\mathrm{distance} = (\mathrm{speed}) * (\mathrm{time})\). Calculate these distances for the times \(t_1\) and \(t_2\) determined in Step 2, which gives \(\mathrm{distance} = 134.67 * 0.35 = 47.14 \mathrm{ft}\) and \(\mathrm{distance} = 134.67 * 0.60 = 80.80 \mathrm{ft}\).

Final Verification

Now, compare these distances to the given horizontal distance of \(60.5 \mathrm{ft}\). The fastball must travel this distance before it hits the ground. Notice that \(60.5 \mathrm{ft}\) falls between \(47.14 \mathrm{ft}\) and \(80.80 \mathrm{ft}\), so the ball does hit inside the strike zone.

Key concepts

Kinematic Equations

Kinematic equations are vital tools in physics when analyzing the motion of objects under constant acceleration. These equations relate an object's initial velocity, final velocity, acceleration, time, and displacement. For projectile motion, we can split the motion into vertical and horizontal components. In the vertical direction, acceleration due to gravity acts on the object causing it to free fall, while in the horizontal direction, the object usually moves at a constant velocity because there are no horizontal forces acting (assuming air resistance is negligible).

Using kinematic equations, we can determine how long a horizontally projected object will remain airborne before it hits the ground and how far it will travel horizontally during this time. The most frequently used kinematic equations in projectile motion are:

• The equation for final velocity: \( v = u + at \)
• The equation for displacement: \( s = ut + \frac{1}{2}at^2 \)
• The equation for the displacement in terms of velocity: \( s = \frac{v^2 - u^2}{2a} \)

In our exercise, we use the second equation for the vertical motion to compute the time for the ball to hit the ground.

Free Fall Acceleration

Free fall acceleration, also known as the acceleration due to gravity, is denoted by the symbol \(g\) and is the acceleration experienced by an object solely under the influence of gravity. It is approximately \(9.8 \text{ m/s}^2\) or \(32.2 \text{ ft/s}^2\), depending on whether you are using metric or imperial units. This acceleration acts on the object in the downward vertical direction regardless of the object's horizontal motion.

In horizontal projectile motion, the only acceleration the object experiences is due to gravity, as there are no forces pushing or pulling it horizontally. Thus, the free fall acceleration is crucial for calculating the time it takes for an object to fall to the ground and to determine its final vertical velocity upon impact.

Unit Conversion

Precise unit conversion is fundamental in physics problems to ensure that measurements are consistent and calculations are correct. This often involves converting from one system of units to another, such as from the imperial system to the metric system, or vice versa. For consistency and to apply kinematic equations correctly, the units of all quantities involved in the equations must be compatible.

In our original exercise, the conversion of speed from miles per hour to feet per second is essential. This is achieved through the use of conversion factors, which are ratios that equal one, multipling the quantity to be converted. Remember that when converting units of speed, you will need both the conversion between miles and feet, and the conversion between hours and seconds.

• To convert speed: \(1 \text{ mi/h} = \frac{5280 \text{ ft}}{3600 \text{ s}}\)
• To convert distance: \(1 \text{ mi} = 5280 \text{ ft}\)
• To convert time: \(1 \text{ h} = 3600 \text{ s}\)

Horizontal Projectile Motion

Horizontal projectile motion refers to the motion of an object that is projected horizontally and moves under the influence of gravity without any initial vertical velocity. The key points in such problems are that horizontally, the velocity remains constant, while vertically, the object accelerates due to gravity with no initial vertical velocity component.

To analyze horizontal projectile motion, we calculate the time an object spends in the air by focusing on the vertical component of the motion. Then, using this time, we compute the horizontal distance the object travels, knowing its horizontal velocity remains unchanged throughout its flight.In our scenario, the baseball is projected horizontally, so we use the time it takes for the ball to fall from the pitcher's hand to determine if it will reach the catcher's mitt within the strike zone.