Question

A baseball is hit at Fenway Park in Boston at a point 0.762 mabove home plate with an initial velocity of 33.53 m/sdirected $\mathbf{55}\mathbf{.}\mathbf{0}\mathbf{°}$above the horizontal. The ball is observed to clear the 11.28-m-high wall in left field (known as the “green monster”)5.00 s after it is hit, at a point just inside the left-field foulline pole. Find (a) the horizontal distance down the left-field foul line from home plate to the wall; (b) the vertical distance by which the ball clears the wall; (c) the horizontal and vertical displacements of the ball with respect to home plate 0.500 sbefore it clears the wall.

Short answer

a) The horizontal distance down the left-field foul line from home plate top of the wall is 96.2 m .

b) The vertical distance by which the ball clears the wall is 4.33 m .

c) The horizontal and vertical displacements of the ball with respect to the home plate are 86.5 m, and 25.12 m respectively.

Step-by-step solution

The given data

• Initial velocity of the ball, $V_0=33.53\mathrm{m}/\mathrm{s}$.
• Angle of the projectile, $\theta =55.0^0$.
• Height of wall, $h=11.28\mathrm{m}$.
• Time taken to clear the wall, $t=5.00\mathrm{s}$.

Understanding the concept of projectile motion

When the particle is launched with some velocity from the ground, its motion is called projectile motion. The path of the particle is calculated by using the set of kinematic equations. For the calculations, the acceleration in the horizontal direction is assumed to be zero, and the upward direction is assumed to be positive. The vertical acceleration is the gravitational acceleration.

Formulae:

The first kinematic equation of motion,$V_f=V_0+at$ (i)

The second kinematic equation of motion,$∆x=V_{0}t+\left(\frac{1}{2}\right)\times a\times t^2$ (ii)

The third kinematic equation of motion,${V_f}^2={V_0}^2+2\times a\times ∆x$ (iii)

Here, $V_0$is the initial velocity,$v_f$ is the final velocity,$a$ is acceleration,$t$ is time, and $∆x$is the displacement.

a) Calculations for the horizontal distance down the left-field foul line from home plate top of the wall

We know that horizontal motion is with constant speed, so$a_x=0$

Components of initial velocity:

$$\begin{gathered} V_{0x}=V_0\times \cos \left(\theta \right) \\ =33.53\times \cos \left(55°\right) \\ =19.23\mathrm{m}/\mathrm{s} \end{gathered}$$

$$\begin{gathered} V_{0y}=V_0\times \cos \left(\theta \right) \\ =33.53\times \sin \left(55°\right) \\ =27.47\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence horizontal distance covered in 5.0 s is given using equation (ii). Substitute the values in the equation (ii) to find the values of x-components.

$$\begin{gathered} x=19.23\times 5+0 \\ =96.15\mathrm{m} \\ \approx 96.2\mathrm{m} \end{gathered}$$

Hence, the value of the horizontal component is $96.2\mathrm{m}$.

b) Calculation for the vertical distance by which the ball clears the wall

Hence vertical distance covered in 5.0s is given using equation (ii). Substitute the values in the equation (ii) to find the values of y-components as:

$$\begin{gathered} y=0.762\mathrm{m}+\left(27.47\mathrm{m}/\mathrm{s}\times 5\mathrm{s}\right)+\left(\frac{1}{2}\right)\times \left(-9.8\mathrm{m}/{\mathrm{s}}^2\right)\times {\left(5\mathrm{s}\right)}^2 \\ =15.612\mathrm{m} \end{gathered}$$

But the vertical distance by which the ball clears the wall is given as:

$$\begin{gathered} d=15.612\mathrm{m}-11.28\mathrm{m} \\ =4.33\mathrm{m} \end{gathered}$$

Hence, the value of the vertical distance is $4.33\mathrm{m}$.

c) Calculation for the horizontal and vertical displacements of the ball with respect to home plate

Now for time,

$$\begin{gathered} t\text{'}=5\mathrm{s}-0.5\mathrm{s} \\ =4.5\mathrm{s} \end{gathered}$$

The horizontal displacement is calculated by substituting the values of initial velocity and time in equation (ii).

$$\begin{gathered} ∆x\text{'}=V_{0x}\times t\text{'} \\ =19.23\mathrm{m}/\mathrm{s}\times 4.5\mathrm{s} \\ =86.5\mathrm{m} \end{gathered}$$

The vertical displacement (y) is found using equation (ii). Substitute the values in the equation (ii).

$$\begin{gathered} y-0.762\mathrm{m}=\left(27.47\mathrm{m}/\mathrm{s}\times 4.5\mathrm{s}\right)+\left(\frac{1}{2}\right)\left(-9.8\mathrm{m}/{\mathrm{s}}^2\right)\times {\left(4.5\mathrm{s}\right)}^2 \\ \mathrm{y}=25.152\mathrm{m} \\ \approx 25.15\mathrm{m} \end{gathered}$$

Hence, the horizontal and vertical displacements of the ball with respect to home plate are 86.5 m, and 25.15 m respectively.