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Chapter 3: Problem 34

MMH On a distant planet, golf is just as popular as it is on earth. A golfer tees off and drives the ball 3.5 times as far as he would have on earth, given the same initial velocities on both planets. The ball is launched at a speed of 45 m/s at an angle of 29 above the horizontal. When the ball lands, it is at the same level as the tee. On the distant planet, what are (a) the maximum height and (b) the range of the ball?

Short Answer

Expert verified
The maximum height on the distant planet is approximately 77.56 meters, and the range is approximately 632.73 meters.

Step by step solution

01

Determine Earth's Range Formula

The range of a projectile launched with speed \( v_0 \) at an angle \( \theta \) on Earth is given by the formula: \( R = \frac{v_0^2 \sin(2\theta)}{g} \). Here, \( g \) is the acceleration due to gravity on Earth, which is approximately \( 9.81 \, \text{m/s}^2 \).
02

Calculate Earth's Range

Substitute the given values into the Earth's range formula: \( v_0 = 45 \, \text{m/s} \) and \( \theta = 29^\circ \).\[ R = \frac{(45)^2 \sin(2 \times 29^\circ)}{9.81} \]Simplify to obtain Earth's range: \[ R_{\text{Earth}} \approx 180.78 \, \text{meters} \].
03

Determine Range on Distant Planet

The range on the distant planet is 3.5 times that on Earth. Thus, \( R_{\text{planet}} = 3.5 \times 180.78 \).Calculate to find: \[ R_{\text{planet}} \approx 632.73 \, \text{meters} \].
04

Determine Maximum Height Formula

The maximum height of a projectile is given by the formula: \( H = \frac{v_0^2 \sin^2(\theta)}{2g} \). Use this formula to first find the height on Earth.
05

Calculate Earth's Maximum Height

Using the maximum height formula and substituting the values \( v_0 = 45 \, \text{m/s} \) and \( \theta = 29^\circ \),\[ H = \frac{(45)^2 \sin^2(29^\circ)}{2 \times 9.81} \]Simplify to calculate: \[ H_{\text{Earth}} \approx 22.16 \, \text{meters} \].
06

Understanding Gravity Impact on Height

Although the gravity is not directly involved in the range relation, the maximum height on a planet would be adjusted proportionally if the range is scaled by 3.5 due to different gravity.
07

Calculate Maximum Height on Distant Planet

Since the range is scaled by 3.5 times, assume gravity is inversely affecting height as a result. Thus, adjust the height by similar scaling: \( H_{\text{planet}} \approx 3.5 \times 22.16 \).Calculate to find: \[ H_{\text{planet}} \approx 77.56 \, \text{meters} \].

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Key Concepts

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

Projectile Range Calculation
In projectile motion, the range of the projectile is crucial to understanding how far it will travel horizontally. The formula used to calculate the range on Earth is given by: \[ R = \frac{v_0^2 \sin(2\theta)}{g} \]where:
  • \( v_0 \) is the initial velocity of the projectile,
  • \( \theta \) is the launch angle,
  • \( g \) is the acceleration due to gravity, approximately 9.81 \( \text{m/s}^2 \) on Earth.
By substituting the launch speed and angle into this formula, we can calculate the range for different conditions and compare them, as in the problem where the range on another planet is much larger than on Earth.
Maximum Height of a Projectile
The maximum height achieved by a projectile is a point where its vertical component of velocity becomes zero before it starts descending. This height can also be calculated using the initial velocity and launch angle with the formula:\[ H = \frac{v_0^2 \sin^2(\theta)}{2g} \]As with the range, changes in gravity affect the maximum height. This means comparing heights on different planets requires considering how velocity and angle play out under different gravitational conditions.
Gravity on Other Planets
Gravity is a force that varies from planet to planet, affecting how objects behave when launched into the air. It is significantly stronger on a planet with higher gravity compared to Earth. In this exercise, the planet's gravity allows a projectile to reach a range 3.5 times longer than on Earth. This means that the force pulling the projectile back to the ground is different, yet proportional calculations show altered projectile behavior. Understanding gravity's impact is crucial for predicting how a projectile will behave when not on Earth.
Projectile Launch Angle
The angle at which a projectile is launched plays a crucial role in determining its range and height. Launch angles that are too steep or too shallow will lessen the range. The optimal angle for maximum range theoretically under uniform conditions is 45 degrees. However, changes in gravity or surface level may impact what angle is actually optimal. In the problem provided, a launch angle of 29 degrees is used, which may be suitable for the conditions on that particular planet.
Physics Problem Solving Steps
Solving physics problems efficiently involves a structured approach:
  • Identify and clearly state the known variables, such as initial velocity and angle of launch.
  • Select the right equations associated with projectile motion; for range and height, these are \( R = \frac{v_0^2 \sin(2\theta)}{g} \) and \( H = \frac{v_0^2 \sin^2(\theta)}{2g} \).
  • Perform the calculations in a sequenced manner, substituting values accurately.
  • Understand each step to ensure that the calculations are consistent with observed phenomena.
This methodical process helps in drawing correct conclusions about the physics of the situation, like in the given scenario where different gravitational conditions were considered.

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Most popular questions from this chapter

A Coast Guard ship is traveling at a constant velocity of 4.20 m/s, due east, relative to the water. On his radar screen the navigator detects an object that is moving at a constant velocity. The object is located at a distance of 2310 m with respect to the ship, in a direction 32.0 south of east. Six minutes later, he notes that the object鈥檚 position relative to the ship has changed to 1120 m, 57.0 south of west. What are the magnitude and direction of the velocity of the object relative to the water? Express the direction as an angle with respect to due west.

A golfer hits a shot to a green that is elevated 3.0 m above the point where the ball is struck. The ball leaves the club at a speed of 14.0 m/s at an angle of 40.0 above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands.

SSM A swimmer, capable of swimming at a speed of 1.4 m/s in still water (i.e., the swimmer can swim with a speed of 1.4 m/s relative to the water), starts to swim directly across a 2.8-km-wide river. However, the current is 0.91 m/s, and it carries the swimmer downstream. (a) How long does it take the swimmer to cross the river? (b) How far downstream will the swimmer be upon reaching the other side of the river?

In the absence of air resistance, a projectile is launched from and returns to ground level. It follows a trajectory similar to that shown in Figure 3.10 and has a range of 23 m. Suppose the launch speed is doubled, and the projectile is fired at the same angle above the ground. What is the new range?

A skateboarder shoots off a ramp with a velocity of 6.6 m/s, directed at an angle of 58 above the horizontal. The end of the ramp is 1.2 m above the ground. Let the x axis be parallel to the ground, the y direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp. (a) How high above the ground is the highest point that the skateboarder reaches? (b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?
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