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Magazine Chapter 3: Problem 36
A certain juggler usually tosses balls vertically to a height \(H\). To what height must they be tossed if they are to spend twice as much time in the air?
Short Answer
Expert verified
The height must be 4H.
Step by step solution
01
- Understand the basics
First, let's recall that the time a projectile spends in the air when tossed vertically is determined by the equation for the time of flight. The equation is given by \( t = \frac{2v_0}{g} \), where \( v_0 \) is the initial velocity and \( g \) is the acceleration due to gravity.
02
- Time of flight for initial height
For the juggler's initial toss, the time of flight when the ball returns to the juggler's hands after reaching height \( H \) can be expressed as \( t = \frac{2v_0}{g} \).
03
- New time of flight
Since the time in the air needs to be doubled, the new time of flight \( t' \) should be \( t' = 2t \). Substituting the initial time, \( t' = 2 \frac{2v_0}{g} = \frac{4v_0}{g} \).
04
- Relationship between flight time and height
The height of the ball is related to the initial velocity by the equation \( H = \frac{v_0^2}{2g} \). To keep the ball in the air for twice the time, we need to find the new height \( H' \) such that \( t' = \frac{4v_0}{g} \).
05
- Finding the new height
To find this new height, we use the fact that the new time is four times the initial time. Using the kinematic equation \( H' = \frac{(2v_0)^2}{2g} \), substituting \( 2v_0 \) for \( v_0' \) since the time has doubled implies the velocity needed to double the time is doubled.
06
- Simplifying the expression
Simplifying the last step, we have \( H' = \frac{4v_0^2}{2g} = 2 \left( \frac{v_0^2}{2g} \right) \). Notice that we have a factor of 4 which shows \( 4H \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
time of flight
The 'time of flight' of a projectile is the total time that the object spends in the air. For an object tossed vertically, the time of flight can be calculated using the formula: \(t = \frac{2v_0}{g}\), where \(v_0\) is the initial velocity and \(g\) is the acceleration due to gravity.
For the juggler's initial toss, the ball returns to his hands after reaching height \(H\). We see that the time of flight directly depends on how fast the ball is thrown upwards.
If the juggler wants the ball to stay in the air for twice as long, he needs to double the initial time spent in the air. Therefore, the new time of flight \(t'\) will be \(2t\).
Understanding this relationship is crucial to analyzing vertical projectile motion accurately.
For the juggler's initial toss, the ball returns to his hands after reaching height \(H\). We see that the time of flight directly depends on how fast the ball is thrown upwards.
If the juggler wants the ball to stay in the air for twice as long, he needs to double the initial time spent in the air. Therefore, the new time of flight \(t'\) will be \(2t\).
Understanding this relationship is crucial to analyzing vertical projectile motion accurately.
initial velocity
The 'initial velocity' \(v_0\) is the speed at which the ball is thrown upwards initially. It's an essential factor in determining how high the ball will go and how long it will stay in the air.
For instance, the formula for the time of flight relies on this initial velocity: \(t = \frac{2v_0}{g}\). The greater the initial velocity, the more time the projectile will spend in the air.
Furthermore, the height \(H\) reached by the ball also depends on this initial velocity, given by the formula: \(H = \frac{v_0^2}{2g}\). To achieve a desired height or time of flight, one must adjust the initial velocity accordingly. When doubling the time in the air, the velocity needs to be increased, which consequently increases the height achieved.
For instance, the formula for the time of flight relies on this initial velocity: \(t = \frac{2v_0}{g}\). The greater the initial velocity, the more time the projectile will spend in the air.
Furthermore, the height \(H\) reached by the ball also depends on this initial velocity, given by the formula: \(H = \frac{v_0^2}{2g}\). To achieve a desired height or time of flight, one must adjust the initial velocity accordingly. When doubling the time in the air, the velocity needs to be increased, which consequently increases the height achieved.
acceleration due to gravity
The 'acceleration due to gravity' (denoted as \(g\)) is a constant that affects the motion of the ball. On Earth, this value is approximately \(9.8 \, m/s^2\). It plays a critical role in the equations governing projectile motion.
For example, both the time of flight and the height reached by the ball are affected by gravity. The formula \(t = \frac{2v_0}{g}\) highlights how a higher gravitational pull would reduce the time of flight for a given initial velocity. Similarly, the height equation \(H = \frac{v_0^2}{2g}\) shows that more gravity means a lower height for the same initial velocity.
This interplay underscores why understanding gravity's impact is fundamental when analyzing any projectile motion scenario.
Therefore, when the juggler wants to keep the ball in the air longer, he must adjust for these gravitational effects by increasing the initial velocity.
For example, both the time of flight and the height reached by the ball are affected by gravity. The formula \(t = \frac{2v_0}{g}\) highlights how a higher gravitational pull would reduce the time of flight for a given initial velocity. Similarly, the height equation \(H = \frac{v_0^2}{2g}\) shows that more gravity means a lower height for the same initial velocity.
This interplay underscores why understanding gravity's impact is fundamental when analyzing any projectile motion scenario.
Therefore, when the juggler wants to keep the ball in the air longer, he must adjust for these gravitational effects by increasing the initial velocity.
