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Chapter 3: Problem 55

The Longest Home Run. According to the Guinness Book of World Records, the longest home run ever measured was hit by Roy "Dizzy" Carlyle in a minor league game. The ball traveled 188 \(\mathrm{m}(618 \mathrm{ft})\) before landing on the ground outside the ballpark. (a) Assuming the ball's initial velocity was \(45^{\circ}\) above the horizotal and ignoring air resistance, what did the initial speed of the ball need to be to produce such a home run if the ball was hit at a point 0.9 \(\mathrm{m}(3.0 \mathrm{ft})\) above ground level? Assume that the ground was perfectly flat. How far would the ball be above a fence 3.0 \(\mathrm{m}\) (10 f) high if the fence was 116 \(\mathrm{m}(380 \mathrm{ft})\) from home plate?

Short Answer

Expert verified
Initial speed: 41.26 m/s, Ball is 0.92 m above the fence at 116 m.

Step by step solution

01

Identify Given Information and Assumptions

Let's first list out the given information and the assumptions. We have:- Horizontal distance traveled: \(d = 188\) m- Launch angle: \(\theta = 45^\circ\)- Height of launch: \(h_0 = 0.9\) m- Height of fence: \(h_f = 3.0\) m- Distance to fence: \(d_f = 116\) mAssume no air resistance and flat ground.
02

Use Projectile Motion Equations

The range (horizontal distance) for projectile motion can be given by the formula:\[ d = \frac{v_0^2 \sin(2\theta)}{g} + \frac{v_0 \cos(\theta) \cdot t_0}{g} \]where \(v_0\) is the initial speed and \(g = 9.81\, \text{m/s}^2\) (acceleration due to gravity).Since the launch and landing heights differ, use:\[ y = v_0 \sin(\theta) \cdot t - \frac{1}{2} gt^2 + h_0 = 0 \] for the range equation, adapting it for the initial height.
03

Find Time to Reach Maximum Height

To find the time at which the projectile reaches maximum height (using vertical motion, \( t_0 \)), we can use:\[ v_y = v_0 \sin(\theta) - gt_0 = 0 \]\[ t_0 = \frac{v_0 \sin(\theta)}{g} \]
04

Solve for Initial Velocity

Plug the expression for time back into the range equation and solve for the initial velocity.Substitute for total time of flight which is twice the ascent time starting from y=0,\[ t = \frac{2v_0 \sin(\theta)}{g} \]Using the range equation:\[ 188 = v_0^2 \sin(90^\circ)/g + \frac{0.9}{v_0 \cos(\theta)} \]Solve this for \(v_0\) which results in:\[ v_0 \approx 41.26 \, \text{m/s} \]
05

Calculate Height Over the Fence at 116 m

To find the height at a given horizontal distance, plug into the trajectory equation:\[ y = x \tan(\theta) - \frac{gx^2}{2(v_0 \cos(\theta))^2} + h_0 \]Use \( x = 116 \) meters:\[ y = 116 \tan(45^\circ) - \frac{9.81 \times 116^2}{2(v_0 \cos(45^\circ))^2} + 0.9 \]\[ y \approx 3.92 \, \text{m} \]
06

Compare Height to Fence

The calculated height at 116 meters is approximately 3.92 meters. The fence is 3.0 meters high, therefore:\[ y_{above\; fence} = 3.92 - 3.0 = 0.92\, \text{m} \]The ball travels 0.92 meters above the fence.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Physics Problem Solving
Physics problem solving, particularly with projectile motion, involves a structured approach to breaking down complex problems into manageable steps.
  • Start by identifying the knowns and unknowns in the problem. For the longest home run case, we know the horizontal distance, launch angle, launch height, and some features of the environment, such as flat ground and negligible air resistance.
  • Make simplifying assumptions, which are crucial. They help reduce complexity. Here, ignoring air resistance simplifies calculations significantly.
  • Choose appropriate equations. In projectile motion, we often use kinematic equations to relate variables like initial velocity, angle, time, and distance.
Breaking down the problem clarifies each step needed to solve the equation. Remember, tackling physics problems is as much about logic as it is about applying formulas.
Projectile Trajectory
Understanding the projectile trajectory involves knowing how a projectile travels through space. This path is typically parabolic. In the longest home run problem, this means:
  • The ball follows a symmetric arc, rising to a peak before descending to the ground.
  • This path is influenced by gravity, which acts downward throughout the flight. Thus, the trajectory can be broken into two components: horizontal and vertical motion.
  • Horizontal motion occurs at a constant speed (if air resistance is ignored). Conversely, vertical motion is affected by gravity, slowing the ascent and accelerating the descent.
When you break down the motion in this way, you can apply the corresponding kinematic equations separately to each component. This separation allows for clear calculation of variables such as range, maximum height, and time of flight.
Initial Velocity Calculation
Finding the initial velocity is key in problems like the longest home run. You need to calculate how fast the ball has to travel at a given angle to achieve a certain range. Here's a simplified process for initial velocity calculation:
  • Use the known range and angle to derive the initial velocity. The angle here is critical because it affects both how high and how far the projectile travels.
  • Apply the range equation, which connects these variables: \[d = \frac{v_0^2 \sin(2\theta)}{g}\]
  • Understand that this equation comes from recognizing that the projectile's path involves both horizontal and vertical motion working together.
In the home run problem, solving the equation for initial velocity lets you predict exactly how powerful a hit is needed to send the ball that far.
Horizontal Range Equation
The horizontal range equation is a key tool in analyzing projectile motion. It allows you to calculate how far a projectile will travel horizontally before hitting the ground. This equation is:\[d = \frac{v_0^2 \sin(2\theta)}{g}\]Important points about this equation include:
  • It's derived considering that maximum range is achieved at a launch angle of 45掳, assuming level launch and landing heights.
  • Variations in launch or landing height, like our home run from a mound, require modifications of the formula for accuracy.
  • Using this equation can help predict outcomes for different initial velocities and angles, allowing you to optimize these parameters for desired results.
Understanding this equation provides insights into how factors like launch speed and angle influence the distance a projectile will cover, exemplified by calculating the initial speed of Roy "Dizzy" Carlyle's famous homerun.

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Most popular questions from this chapter

A projectile is launched with speed \(v_{0}\) at an angle \(\alpha_{0}\) above the horizontal. The launch point is a height \(h\) above the ground. (a) Show that if air resistance is ignored, the horizontal distance that the projectile travels before striking the ground is $$ x=\frac{v_{0} \cos \alpha_{0}}{g}\left(v_{0} \sin \alpha_{0}+\sqrt{v_{0}^{2} \sin ^{2} \alpha_{0}+2 g h}\right) $$ Verify that if the launch point is at ground level so that \(h=0\) , this is equal to the horizontal range \(R\) found in Example \(3.8 .\) (b) For the case where \(v_{0}=10 \mathrm{m} / \mathrm{s}\) and \(h=5.0 \mathrm{m},\) graph \(x\) as a function of launch angle \(\alpha_{0}\) for values of \(\alpha_{0}\) from \(0^{\circ}\) to \(90^{\circ} .\) Your graph should show that \(x\) is zero if \(\alpha_{0}=90^{\circ},\) but \(x\) is nonzero if \(\alpha_{0}=0\) ; explain why this is so. (c) We saw in Example 3.8 that for a projectile that lands at the same height from which it is launched, the horizontal range is maximum for \(\alpha_{0}=45^{\circ} .\) For the case graphed in part \((b),\) is the angle for maximum horizontal distance equal to, less than, or greater than \(45^{\circ} 7\) (This is a general result for the situation where a projectile is launched from a point higher than where it lands.)

On your first day at work for an appliance manufacturer, you are told to figure out what to do to the period of rotation during a washer spin cycle to triple the centripetal accelerntion. You inpress your boss by answering immediately. What do you tell her?

On the Flying Trapeze. new circus act is called the Texas Tumblers. Lovely Mary Belle swings from a trapeze, projects herself at an angle of \(53^{\circ},\) and is supposed to be canght by Joe Bob, whose hands are 6.1 \(\mathrm{m}\) above and 8.2 \(\mathrm{m}\) horizontally from her launch point (Fig. 3.50 ) You can ignore air resistance. (a) What initial speed \(v_{0}\) must Mary Belle have just to reach Joe Bob? (b) For the initial speed calculated in part (a), what are the magnitude and direction of her velocity when Mary Belle reaches Joe Bob? (c) Assuming that Mary Belle has the initial speed calculated in part (a), draw \(x-t, y-t, v_{x}-t,\) and \(v_{y}-t\) graphs showing the motion of both tumblers. Your graphs should show the motion up until the point where Mary Belle reaches Joe Bob. (d) The night of their debut performance, Joe Bob misses her completely as she flies past. How far horizontally does Mary Belle travel, from her initial launch point, before landing in the safety net 8.6 \(\mathrm{m}\) below her starting point?

A web page designer creates an animation in which a dot on a computer screen has a position of \(\vec{r}=[4.0 \mathrm{cm}+\) \(\left(2.5 \mathrm{cm} / \mathrm{s}^{2}\right) t^{2} ] \hat{\imath}+(5.0 \mathrm{cm} / \mathrm{s}) \hat{t}\) . (a) Find the magnitude and direction of the dot's average velocity between \(t=0\) and \(t=2.0 \mathrm{s} .\) (b) Find the magnitude and direction of the instantaneous velocity at \(t=0, t=1.0 \mathrm{s},\) and \(t=2.0 \mathrm{s}\) . (c) Sketch the dot's trajectory from \(t=0\) to \(t=2.0 \mathrm{s}\) , and show the velocities calculated in part \((\mathrm{b})\) .

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