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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 quarterback claims that he can throw the football a horizontal distance of 183 m (200 yd). Furthermore, he claims that he can do this by launching the ball at the relatively low angle of 30.0 above the horizontal. To evaluate this claim, determine the speed with which this quarterback must throw the ball. Assume that the ball is launched and caught at the same vertical level and that air resistance can be ignored. For comparison, a baseball pitcher who can accurately throw a fastball at 45 m/s (100 mph) would be considered exceptional.

SSM An eagle is flying horizontally at 6.0 m/s with a fish in its claws. It accidentally drops the fish. (a) How much time passes before the fish鈥檚 speed doubles? (b) How much additional time would be required for the fish鈥檚 speed to double again?

MMH After leaving the end of a ski ramp, a ski jumper lands downhill at a point that is displaced 51.0 m horizontally from the end of the ramp. His velocity, just before landing, is 23.0 m/s and points in a diretion 43.0-below the horizontal. Neglecting air resistance and any lift he experiences while airborne, find his initial velocity (magnitude and direction) when he left the end of the ramp. Express the direction as an angle relative to the horizontal.

In the javelin throw at a track-and-field event, the javelin is launched at a speed of \(29 \mathrm{~m} / \mathrm{s}\) at an angle of \(36^{\circ}\) above the horizontal. As the javelin travels upward, its velocity points above the horizontal at an angle that decreases as time passes. How much time is required for the angle to be reduced from \(36^{\circ}\) at launch to \(18^{\circ} ?\)

The lob in tennis is an effective tactic when your opponent is near the net. It consists of lofting the ball over his head, forcing him to move quickly away from the net (see the drawing). Suppose that you lob the ball with an initial speed of 15.0 m/s, at an angle of 50.0 above the horizontal. At this instant your opponent is 10.0 m away from the ball. He begins moving away from you 0.30 s later, hoping to reach the ball and hit it back at the moment that it is 2.10 m above its launch point. With what minimum average speed must he move? (Ignore the fact that he can stretch, so that his racket can reach the ball before he does.)
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