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(a) During a tennis match, a player serves at \(23.6 \mathrm{~m} / \mathrm{s}\) (as recorded by radar gun), with the ball leaving the racquet \(2.37 \mathrm{~m}\) above the court surface, horizontally. By how much does the ball clear the net, which is \(12 \mathrm{~m}\) away and \(0.90 \mathrm{~m}\) high? (b) Suppose the player serves the ball as before except that the ball leaves the racquet at \(5.0^{\circ}\) below the horizontal. Does the ball clear the net now?

Short Answer

Expert verified
In part (a), the ball clears the net. However, in part (b), the ball does not clear the net.

Step by step solution

01

Find Time to Reach the Net in Part (a)

The time it takes for the ball to reach the net can be calculated using the formula for uniform motion: \(d = vt\), where \(d\) is distance (12 m), \(v\) is velocity (23.6 m/s), and \(t\) is time. Solve this equation for \(t\): \(t = d/v\).
02

Calculate Descent of the Ball in Part (a)

Now we find out how much the ball would have fallen during this time. We use the equation of motion: \(d = ut +(1/2)gt^2\), where \(u\) is initial vertical velocity (0), \(g\) is acceleration due to gravity (9.8 m/s²), \(t\) is time from Step 1, and \(d\) is the distance (descent). As the initial vertical velocity (\(u\)) of the ball is 0, this leaves us with the equation \(d =(1/2)gt^2\). From this, we find the total descent of the ball.
03

Clearance of the Ball Over the Net in Part (a)

Now, to find out whether the ball clears the net or not, we subtract the height of the net (0.90 m) from the initial height of the ball (2.37 m) and the descent calculated in the previous step. If the result is positive, the ball clears the net, else it does not.
04

Find Time to Reach the Net in Part (b)

Using the same method as in Step 1, we can find the time it takes for the ball to reach the net if served at an angle. We use the same formula but the horizontal velocity (v) now becomes the horizontal component of the total velocity which is \(23.6 \cos (5.0) \) m/s. So, \(t = d/v\).
05

Calculate Descent of the Ball in Part (b)

In this case, the initial vertical velocity (\(u\)) of the ball is \(23.6 \sin(5.0)\) m/s (downward). Using the case (b) time found from Step 4 in the earlier equation of motion, we can find the total descent of the ball.
06

Clearance of the Ball Over the Net in Part (b)

We use the same way as in Step 3 to find out whether the ball clears the net or not. We subtract the height of the net (0.90 m) from the initial height of the ball (2.37 m) and the descent calculated in the previous step. If the result is positive, the ball clears the net, else it does not.

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Key Concepts

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

Uniform Motion
Uniform motion is a type of motion where an object travels in a straight line with a constant speed. This concept is a cornerstone in physics because it simplifies understanding the movement of objects. When dealing with uniform motion, you often use the fundamental formula:
  • Distance, \(d\) = Speed, \(v\) × Time, \(t\)
If you know any two of these quantities, you can easily find the third. In the context of a tennis serve, the uniform motion helps us figure out how long the ball takes to travel a certain horizontal distance—like reaching the net.
This is crucial in analyzing projectile motion where vertical and horizontal components are treated separately.
The horizontal motion, because it’s not influenced by external forces like gravity in the ideal case, is considered uniform.
Equation of Motion
Equations of motion describe how an object moves under the influence of forces. For a tennis ball served horizontally, the equation of motion we primarily use is:
  • \(d = ut + \frac{1}{2}gt^2\)
Here, \(d\) is the vertical distance fallen, \(u\) is initial vertical velocity, \(g\) is the acceleration due to gravity, and \(t\) is time.
In the case of a horizontal serve, the initial vertical velocity \(u\) is 0, transforming the equation to
\(d = \frac{1}{2}gt^2\).
We use this to determine how far the ball falls while traveling to the net.
For shots with an angle, initial vertical velocity is calculated using trigonometric components of the total velocity.
Projectile Angle
When a ball is served at an angle, it follows a projectile path. The projectile angle, in this scenario, affects both the horizontal and vertical components of motion.
  • The horizontal component is given by: \(v_x = v \cos(\theta)\)
  • The vertical component is calculated by: \(v_y = v \sin(\theta)\)
Where \(v\) is the initial speed and \(\theta\) is the serve angle.
Each of these components influences how the ball moves toward the net.
Serves at angles have a more complex trajectory than horizontal serves, as they combine both falling due to gravity and the directional launch speed.
Understanding how to decompose the velocity into components lets you predict where the ball will land, and helps in determining its net clearance.
Gravity
Gravity is the force that pulls objects towards the Earth, affecting all moving objects. It accelerates objects downward at \(9.8 \text{ m/s}^2\). Even during something as simple as a tennis serve, gravity plays a critical role.
It's what causes the ball to start falling the moment it leaves the racquet.
  • Gravity’s consistent pull allows us to use the equation of motion efficiently to predict vertical descent.
  • In horizontal motion, gravity doesn’t affect speed, but it significantly impacts the vertical path.
In projectile motion, the constant acceleration due to gravity ensures that no matter how fast the horizontal velocity is, eventually the ball will fall to the ground.
Understanding gravity’s impact allows players to adjust their techniques to optimize their serves and match strategies. It’s also a reminder that in any motion analysis, gravity cannot be neglected.

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Most popular questions from this chapter

A particle leaves the origin at \(t=0\) with an initial velocity \(\overrightarrow{\mathbf{v}}_{0}=(3.6 \mathrm{~m} / \mathrm{s}) \hat{\mathbf{i}}\). It experiences a constant acceleration \(\overrightarrow{\mathbf{a}}=-\left(1.2 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathbf{i}}-\left(1.4 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathbf{j}} .(a)\) At what time does the particle reach its maximum \(x\) coordinate? \((b)\) What is the velocity of the particle at this time? \((c)\) Where is the particle at this time?

A ball rolls off the edge of a horizontal tabletop, \(4.23 \mathrm{ft}\) high. It strikes the floor at a point \(5.11 \mathrm{ft}\) horizontally away from the edge of the table. ( \(a\) ) For how long was the ball in the air? (b) What was its speed at the instant it left the table?

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 \(3.60 \mathrm{ft}\).

A batter hits a pitched ball at a height \(4.0 \mathrm{ft}\) above the ground so that its angle of projection is \(45^{\circ}\) and its horizontal range is \(350 \mathrm{ft}\). The ball travels down the left field line where a 24 -fthigh fence is located \(320 \mathrm{ft}\) from home plate. Will the ball clear the fence? If so, by how much?

A particle moves so that its position as a function of time is $$ \overrightarrow{\mathbf{r}}(t)=A \hat{\mathbf{i}}+B t^{2} \hat{\mathbf{j}}+C t \hat{\mathbf{k}} $$ where \(A=1.0 \mathrm{~m}, B=4.0 \mathrm{~m} / \mathrm{s}^{2}\), and \(C=1.0 \mathrm{~m} / \mathrm{s}\). Write ex- pressions for \((a)\) its velocity and \((b)\) its acceleration as functions of time. \((c)\) What is the shape of the particle's trajectory?

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