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Inside a star ship at rest on the earth, a ball rolls off the top of a horizontal table and lands a distance \(D\) from the foot of the table. This star ship now lands on the unexplored Planet \(X\) . The commander, Captain Curious, rolls the same ball off the same table with the same initial speed as on earth and finds that it lands a distance 2.76\(D\) from the foot of the table. What is the acceleration due to gravity on Planet \(\mathrm{X}\) ?

Short Answer

Expert verified
The acceleration due to gravity on Planet X is approximately 1.287 m/s².

Step by step solution

01

Find time of fall on Earth

On Earth, the ball rolls off from the table and follows a projectile motion. First, we calculate the time it takes to reach the ground. Assuming the ball falls freely from the level of the table, its initial vertical speed is 0.Using the equation for free fall: \[ h = \frac{1}{2} g t^2 \]Where:- \( h \) is the height of the table (unknown),- \( g \) is the acceleration due to gravity on Earth (approximately \( 9.81 \, m/s^2 \)),- \( t \) is the time taken to hit the ground. We don't know the height or the time directly yet, but we know the time \( t \) required for any horizontal distance \( D = v_0 t \), where \( v_0 \) is the horizontal speed.
02

Relate horizontal motion on Earth to vertical motion

The ball travels the same horizontal distance \( D \) during time \( t \) with speed \( v_0 \), so \( v_0 = \frac{D}{t} \) is constant.Thus, the horizontal motion is: \[ D = v_0 \times t \]Also, using the vertical motion \[ t = \sqrt{\frac{2h}{g}} \]Combine to eliminate \( t \) in terms of \( D \):\[ D = v_0 \times \sqrt{\frac{2h}{g}} \]
03

Determine horizontal distance on Planet X

The same equations apply on Planet X, but now the distance is \( 2.76D \).Hence,\[ 2.76D = v_0 \times \sqrt{\frac{2h}{g_x}} \]where \( g_x \) is the unknown gravitational acceleration on Planet X.
04

Express time \( t_x \) and relations

Using the explored ratio from Step 2, solve the equation using the known ratio of horizontal distances, which will help find \( g_x \):Divide distance equation of Earth into equation of Planet X:\[ \frac{2.76D}{D} = \frac{\sqrt{2h/g_x}}{\sqrt{2h/g}} \]This simplifies to:\[ 2.76 = \sqrt{\frac{9.81}{g_x}} \]
05

Solve for \( g_x \)

Squaring both sides of the above equation:\[ 2.76^2 = \frac{9.81}{g_x} \]Calculate \( 2.76^2 = 7.6176 \):\[ 7.6176 = \frac{9.81}{g_x} \]Hence,\[ g_x = \frac{9.81}{7.6176} \approx 1.287 \, \text{m/s}^2 \]
06

Conclude with acceleration on Planet X

The calculations show that the gravitational acceleration \( g_x \) on Planet X is approximately \( 1.287 \, m/s^2 \).

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Key Concepts

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

Acceleration Due to Gravity
When discussing projectile motion, one of the key factors involved is the acceleration due to gravity. On Earth, this is approximately constant at \(9.81 \, \text{m/s}^2\). This value represents how quickly an object will accelerate downwards when in free fall, regardless of its horizontal movement.
The acceleration due to gravity (\(g\)) is crucial in determining how long an object takes to hit the ground from a height. In the original exercise, the differing accelerations on Earth compared to Planet \(X\) affected how far the ball traveled horizontally given the same initial conditions.
This concept explains why, if gravity is weaker on Planet \(X\), as calculated to be approximately \(1.287 \, \text{m/s}^2\), the ball travels a greater horizontal distance, because it falls more slowly, staying in motion for longer before hitting the ground.
Free Fall Equations
Free fall equations are fundamental in calculating how long an object in projectile motion takes to fall from a certain height. They help relate the vertical motion to time by considering gravity's effect on the object.
The main equation is \( h = \frac{1}{2} g t^2 \), where \( h \) is the height from which the object is dropped, \( g \) is the gravitational acceleration, and \( t \) is the time taken to fall.
  • This equation assumes the initial vertical speed is zero when an object falls from rest.
  • The equation helps determine the time \( t \), which is crucial for calculating how far an object can travel horizontally before hitting the ground.
In the exercise, the height \( h \) was unknown, but by using the free fall equations, we could relate \( h \) to time and eventually solve for the horizontal distance traveled under different gravitational forces.
Horizontal Motion
In projectile motion, horizontal motion is considered separately from vertical motion. Although the object is affected by gravity, its horizontal speed remains constant (as long as air resistance is ignored).
For horizontal motion, the formula \( D = v_0 \times t \) describes how distance \( D \) is proportional to the initial horizontal speed \( v_0 \) and the time \( t \) it remains in motion. This equation can be crucial in predicting where an object will land.
  • At the same initial speed, the time period for which the object stays airborne determines how far it travels horizontally.
  • Hence, if the time of flight increases due to lower gravitational acceleration, the horizontal distance traveled increases as well.
In the given exercise, these principles explain why the same ball rolled a further distance on Planet \(X\) than on Earth, owing to the difference in gravity, thereby increasing its in-air time.

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Most popular questions from this chapter

. The nozzle of a fountain jet sits in the center of a circular pool of radius 3.50 \(\mathrm{m}\) . If the nozzle shoots water at an angle of \(65^{\circ}\) , what is the maximum speed of the water at the nozzle that will allow it to land within the pool? (You can ignore air resistance.)

A bottle rocket can shoot its projectile vertically to a height of 25.0 \(\mathrm{m}\) . At what angle should the bottle rocket be fired to reach its maximum horizontal range, and what is that range? (You can ignore air resistance.)

A baseball pitcher throws a fastball horizontally at a speed of 42.0 \(\mathrm{m} / \mathrm{s} .\) Ignoring air resistance, how far does the ball drop between the pitcher's mound and home plate, 60 \(\mathrm{ft} 6\) in away?

An errand of mercy. An airplane is dropping bales of hay to cattle stranded in a blizzard on the Great Plains. The pilot releases the bales at 150 \(\mathrm{m}\) above the level ground when the plane is flying at 75 \(\mathrm{m} / \mathrm{s} 55^{\circ}\) above the horizontal. How far in front of the cattle should the pilot release the hay so that the bales will land at the point where the cattle are stranded?

At an air show, a jet plane has velocity components \(v_{x}=\) 625 \(\mathrm{km} / \mathrm{h}\) and \(v_{y}=415 \mathrm{km} / \mathrm{h}\) at time 3.85 \(\mathrm{s}\) and \(v_{x}=838 \mathrm{km} / \mathrm{h}\) and \(v_{y}=365 \mathrm{km} / \mathrm{h}\) at time 6.52 s. For this time interval, find(a) the \(x\) and \(y\) components of the plane's average acceleration and (b) the magnitude and direction of its average acceleration.

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