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Chapter 4: Problem 131

A golfer tees off from the top of a rise, giving the golf ball an initial velocity of \(43.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(30.0^{\circ}\) above the horizontal. The ball strikes the fairway a horizontal distance of \(180 \mathrm{~m}\) from the tee. Assume the fairway is level. (a) How high is the rise above the fairway? (b) What is the speed of the ball as it strikes the fairway?

Short Answer

Expert verified
(a) The rise is approximately 10.0 m. (b) The ball's speed when it strikes the fairway is about 41.0 m/s.

Step by step solution

01

Break Down Initial Velocity into Components

The initial velocity (\(v_0\)) can be split into horizontal (\(v_{0x}\)) and vertical (\(v_{0y}\)) components using trigonometric functions. \(v_{0x} = v_0 \cos \theta = 43.0 \cos 30.0^{\circ}\) and \(v_{0y} = v_0 \sin \theta = 43.0 \sin 30.0^{\circ}\). Compute these components.
02

Calculate Time of Flight

Assume horizontal motion: \(x = v_{0x} t\), where \(x = 180 \mathrm{~m}\). Solve for \(t\) to find how long the ball is in the air. Use \(t = \frac{x}{v_{0x}}\).
03

Use Vertical Motion Equation

Apply vertical motion equations to find the height of the rise. Use the equation \(y = v_{0y}t + \frac{1}{2}a t^2\), where \(a = -9.8 \mathrm{~m/s}^2\). Solve for \(y\) after substituting \(t\) from Step 2.
04

Calculate the Final Velocity

Determine the final velocity components. Use horizontal component as \(v_{x} = v_{0x}\) and find vertical component with \(v_y = v_{0y} + at\), using \(t\) from Step 2. Find the magnitude of the velocity \(v = \sqrt{v_x^2 + v_y^2}\).
05

Find the Speed When Ball Strikes Fairway

Now that you have both velocity components at impact, find the final speed at which the ball strikes the fairway using \(v = \sqrt{v_x^2 + v_y^2}\).
06

Conclusion

Substituting all known values and performing the calculations will yield the height of the rise (part a) and the speed of the ball (part b).

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Key Concepts

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

Initial Velocity Components
In projectile motion, the initial velocity is often given as a single vector with a magnitude and direction. This vector can be divided into two components: horizontal and vertical. This is important because these components interact differently due to gravity.
  • Horizontal Component (\(v_{0x}\)): This is calculated using the cosine function of the angle of launch. It remains constant because there's typically no horizontal acceleration acting on the object. For example, in our exercise, \(v_{0x} = 43.0 \cos 30.0^{\circ} \approx 37.2 \mathrm{~m/s}\).
  • Vertical Component (\(v_{0y}\)): This component is affected by gravity. It is calculated using the sine function of the launch angle. In our exercise, \(v_{0y} = 43.0 \sin 30.0^{\circ} = 21.5 \mathrm{~m/s}\).
These components allow us to handle the horizontal and vertical motions separately using their respective equations of motion.
Time of Flight
The time of flight in a projectile motion problem refers to the total time the projectile remains in the air. It is determined by considering the horizontal motion.
To calculate the time of flight:
  • Use the formula for horizontal distance: \(x = v_{0x} \, t\).
  • With \(x = 180 \mathrm{~m}\) and \(v_{0x} \approx 37.2 \mathrm{~m/s}\), solve for \(t\):
    \(t = \frac{180 \mathrm{~m}}{37.2 \mathrm{~m/s}} \approx 4.84 \mathrm{~s}\).
This time is crucial for analyzing vertical motion and determining the projectile's position at any point during its flight.
Vertical Motion Equation
Vertical motion in projectile motion is influenced by gravity, requiring a different approach than horizontal motion. To find aspects such as the height from which a projectile was launched, we use the vertical motion equation:
  • The standard equation is \(y = v_{0y}t + \frac{1}{2}a t^2\), where \(a = -9.8 \mathrm{~m/s}^2\) due to gravity.
  • Substitute the known values: \(v_{0y} = 21.5 \mathrm{~m/s}\), \(t = 4.84 \mathrm{~s}\), and \(a = -9.8 \mathrm{~m/s}^2\).
  • Calculate the height: \(y = (21.5 \mathrm{~m/s}) (4.84 \mathrm{~s}) + \frac{1}{2}(-9.8 \mathrm{~m/s}^2)(4.84 \mathrm{~s})^2\).
  • This calculation provides the vertical displacement or the rise height, which is around 25.3 meters above the fairway.
This equation is pivotal in determining vertical differences in projectile problems.
Final Velocity Calculation
Determining the final velocity of a projectile involves both its horizontal and vertical velocity components at the moment of impact.
  • Horizontal Velocity (\(v_x\)): Since horizontal velocity remains unaffected by gravity, it equals the initial horizontal velocity: \(37.2 \mathrm{~m/s}\).
  • Vertical Velocity (\(v_y\)): This changes due to gravity. Use the formula: \(v_y = v_{0y} + at\).
  • Calculate: \(v_y = 21.5 \mathrm{~m/s} + (-9.8 \mathrm{~m/s}^2)(4.84 \mathrm{~s}) \approx -25.8 \mathrm{~m/s}\).
  • Combine the components to find the magnitude of the final velocity: \(v = \sqrt{v_x^2 + v_y^2}\).
  • Substitute the known values: \(v = \sqrt{(37.2)^2 + (-25.8)^2} \approx 45.8 \mathrm{~m/s}\).
This calculation gives the speed at which the ball hits the fairway, providing complete insight into its landing dynamics.

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Most popular questions from this chapter

A proton initially has \(\vec{v}=4.0 \hat{\mathrm{i}}-2.0 \hat{\mathrm{j}}+3.0 \hat{\mathrm{k}}\) and then \(4.0 \mathrm{~s}\) later has \(\vec{v}=-2.0 \hat{\mathrm{i}}-2.0 \hat{\mathrm{j}}+5.0 \hat{\mathrm{k}}\) (in meters per second). For that \(4.0 \mathrm{~s},\) what are \((\mathrm{a})\) the proton's average acceleration \(\vec{a}_{\mathrm{avg}}\) in unitvector notation, (b) the magnitude of \(\vec{a}_{\text {avg }},\) and (c) the angle between \(\vec{a}_{\text {avg }}\) and the positive direction of the \(x\) axis?

A carnival merry-go-round rotates about a vertical axis at a constant rate. A man standing on the edge has a constant speed of \(3.66 \mathrm{~m} / \mathrm{s}\) and a centripetal acceleration \(\vec{a}\) of magnitude \(1.83 \mathrm{~m} / \mathrm{s}^{2}\) Position vector \(\vec{r}\) locates him relative to the rotation axis. (a) What is the magnitude of \(\vec{r} ?\) What is the direction of \(\vec{r}\) when \(\vec{a}\) is directed (b) due east and (c) due south?

A football kicker can give the ball an initial speed of \(25 \mathrm{~m} / \mathrm{s} .\) What are the (a) least and (b) greatest elevation angles at which he can kick the ball to score a field goal from a point \(50 \mathrm{~m}\) in front of goalposts whose horizontal bar is \(3.44 \mathrm{~m}\) above the ground?

A rifle is aimed horizontally at a target \(30 \mathrm{~m}\) away. The bullet hits the target \(1.9 \mathrm{~cm}\) below the aiming point. What are (a) the bullet's time of flight and (b) its speed as it emerges from the rifle?

A woman who can row a boat at \(6.4 \mathrm{~km} / \mathrm{h}\) in still water faces a long, straight river with a width of \(6.4 \mathrm{~km}\) and a current of \(3.2 \mathrm{~km} / \mathrm{h}\). Let i point directly across the river and j point directly downstream. If she rows in a straight line to a point directly opposite her starting position, (a) at what angle to 矛 must she point the boat and (b) how long will she take? (c) How long will she take if, instead, she rows \(3.2 \mathrm{~km}\) down the river and then back to her starting point? (d) How long if she rows \(3.2 \mathrm{~km}\) up the river and then back to her starting point? (e) At what angle to i should she point the boat if she wants to cross the river in the shortest possible time? (f) How long is that shortest time?
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