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Chapter 3: Problem 69

A golf ball with an initial speed of \(50.0 \mathrm{~m} / \mathrm{s}\) lands exactly \(240 \mathrm{~m}\) downrange on a level course. (a) Neglecting air friction, what tus projection angles would achieve this result? (b) What is the maximum height reached by the ball, using the two angles determined in part (a)?

Short Answer

Expert verified
The golf ball can land exactly 240 m away on a level course if it is projected at angles of approximately 16.26 degrees or 73.74 degrees. The maximum heights corresponding to these two angles are approximately 60.8 m and 60.8 m, respectively.

Step by step solution

01

Calculation of projection angles

Use the formula for the range of a projectile which is given by: \[ R = \frac{{v^2}}{g} \sin{2\theta} \]. Given that the Range \( R = 240\,m \), initial speed \( v = 50.0 \,m/s \), and acceleration due to gravity \( g = 9.8 \,m/s^2 \). Rearranging the formula for \( \theta \), we get: \( \theta = \frac{1}{2} \sin^{-1} \left(\frac{R \cdot g}{v^2} \right) \). Solving this equation will give the two projection angles.
02

Calculation of maximum height for each angle

The maximum height \( H \) reached by a projectile is given by: \( H = \frac{v^2 \sin^{2}\theta}{2g} \). Use the two angles determined in the previous step to compute the maximum height for each angle.

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Key Concepts

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

Projection Angle
The projection angle in projectile motion is a crucial factor that influences the path and outcome of a projectile, such as a golf ball. For a given initial speed, different angles can lead to different ranges and trajectories. The key is choosing the right angle to achieve the desired outcome. In the case of a golf ball being hit on a level course:
  • Consider the formula for the range of a projectile: \[ R = \frac{{v^2}}{g} \sin{2\theta} \]Where \(v\) is the initial speed, \(g\) is the acceleration due to gravity, and \(\theta\) is the angle above the horizontal.
  • By rearranging the formula, we solve for \(\theta\), yielding:\( \theta = \frac{1}{2} \sin^{-1} \left(\frac{R \cdot g}{v^2} \right) \)
For the given exercise, different projection angles are possible to achieve the same range of 240 meters. Understanding these calculations allows one to find possible angles which affect the ball's travel path.
Range of a Projectile
The range of a projectile is the total horizontal distance it covers before landing. To maximize this range, an understanding of initial conditions such as speed and angle is essential.
Using the formula mentioned above, the range affected by:
  • Initial speed \(v\), where higher speeds generally result in longer ranges.
  • Projection angle \(\theta\), where specific angles can maximize the distance the projectile travels.
  • Gravity \(g\), which acts to pull the projectile downwards.
In the real world, neglecting air resistance, angles of 45 degrees provide the maximum range for any initial speed. However, in scenarios such as this golf ball problem with a given range, two angles can achieve the desired result due to the trigonometric nature of the sine function.
Maximum Height
The maximum height a projectile reaches is determined by the initial speed and the sine of the projection angle. The formula to find the maximum height (\(H\)) is:\[ H = \frac{v^2 \sin^{2}\theta}{2g} \]
This height depends significantly on:
  • The initial speed \(v\), where faster projectiles can reach greater heights.
  • The angle \(\theta\), which dictates the vertical component of the projectile's velocity.
      • In case of this exercise, after determining the two angles that achieve the desired range, use each angle to calculate the corresponding maximum height. Observing these heights can provide insight into how changing angles affect the projectile's peak.
Physics Problem Solving
Physics problem solving requires a structured approach where understanding the underlying principles is key. When tackling projectile motion problems, such as hitting a golf ball:
  • Start with all known quantities, like initial speed and desired range.
  • Apply relevant equations of motion, such as those for range and maximum height.
  • Simplify and rearrange equations to isolate the unknown variables, like the angle or height.
Practicing these steps helps develop a strong intuition for physics problems, allowing for quick adjustments based on different scenarios or conditions. With each problem, you'll refine your ability to predict and calculate outcomes using foundational physics principles.

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Most popular questions from this chapter

A boat moves through the water of a river at \(10 \mathrm{~m} / \mathrm{s}\) relative to the water, regardless of the boat's direction. If the water in the river is flowing at \(1.5 \mathrm{~m} / \mathrm{s}\), how long does it Lake the boat to make a round trip consisting of a \(900-\mathrm{m}\) displacement downstream followed by a \(300-\mathrm{m}\) displacement upstream?

A golfer takes two putts to get his ball into the hole once he is on the green. The first putt displaces the ball \(6.00 \mathrm{~m}\) east, the second \(5.40 \mathrm{~m}\) south. What displacement would have been needed to get the ball into the hole on the first putt?

An artillery shell is fired with an initial velocity of \(300 \mathrm{~m} / \mathrm{s}\) at \(55.0^{\circ}\) above the horizontal. To clear an avalanche, it explodes on a mountainside \(42.0 \mathrm{~s}\) after firing. What are the \(x\) -and \(y\) -coordinates of the shell where it explodes, relative to its firing point?

A Nordic jumper goes off a ski jump at an angle of \(10.0^{\circ}\) below the horizontal, traveling \(108 \mathrm{~m}\) horizontally and \(55.0 \mathrm{~m}\) vertically before landing. (a) Ignoring friction and aerodynamic effects, calculate the speed needed by the skier on leaving the ramp. (b) Olympic Nordic jumpers can make such jumps with a jump speed of \(29.0 \mathrm{~m} / \mathrm{s}\), which is considerably less than the answer found in part (a). Explain how that is possible.

How long does it take an automobile traveling in the left lane of a highway at \(60.0 \mathrm{~km} / \mathrm{h}\) to overtake (become even with) another car that is traveling in the right lane at \(40.0 \mathrm{~km} / \mathrm{h}\) when the cars' front bumpers are initially \(100 \mathrm{~m}\) apart?
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