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Chapter 4: Problem 6

In quick succession, a large number of balls are throw vertically in such a way that the next ball is thrown ou up when the previous ball is at the maximum height. If the maximum height is \(5 \mathrm{~m}\); then find the number of the throw: up per second \(\left(\mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{-2}\right)\).

Short Answer

Expert verified
1 throw per second.

Step by step solution

01

Determine the Time to Reach Maximum Height

First, let's determine how long it takes for a ball to reach its maximum height of 5 meters. We know that the velocity at the maximum height is 0. We can use the following kinematic equation:\[v^2 = u^2 - 2gh\]where:- \(v = 0\) (final velocity at the maximum height),- \(u\) is the initial velocity,- \(g = 10 \, \text{m/s}^2\),- \(h = 5 \, \text{m}.\)Rearranging the formula to solve for \(u\), we get:\[u^2 = 2 \, g \, h = 2 \cdot 10 \cdot 5 = 100\]So, \(u = 10 \, \text{m/s}\).
02

Calculate Time to Maximum Height

Now that we know the initial velocity \(u = 10 \, \text{m/s}\), we can calculate the time \(t\) taken to reach the maximum height using: \[v = u - gt\]Substituting the known values:\[0 = 10 - 10t\]\[t = 1 \, \text{s}\]Thus, it takes 1 second for a ball to reach the maximum height.
03

Determine Throws per Second

Since each ball takes 1 second to reach its maximum height and the next ball is thrown at the max height of the first, there is a throw made every 1 second.

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Key Concepts

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

Understanding Kinematic Equations
Kinematic equations are essential tools for analyzing motion, especially in projectile motion scenarios. They relate various motion parameters such as velocity, acceleration, time, and displacement. In this exercise, we used one of these kinematic equations to determine the initial velocity needed for a ball to reach a specific height.

The specific kinematic equation used is:
  • \[v^2 = u^2 - 2gh\]
Where:
  • \(v\) is the final velocity,
  • \(u\) is the initial velocity,
  • \(g\) is the gravitational acceleration, and
  • \(h\) is the height reached.
We set \(v = 0\) at maximum height since the ball momentarily stops before descending. By rearranging the kinematic equation, we can solve for the initial speed \(u\). This helps understand how high and how quickly the objects can move under the influence of gravity.
Insights into Vertical Velocity
Vertical velocity is a crucial component when dealing with projectile motion like in our exercise. It determines how quickly an object moves upward until it momentarily stops at its peak. In vertical motion, the velocity decreases as the object rises, due to gravitational pull acting against it.

Initially, we calculated the initial vertical velocity \(u\) for the ball using:
  • \[u^2 = 2gh\]
This gives us an understanding that the velocity is directly related to the height. Furthermore, using the equation:
  • \[v = u - gt\]
we were able to determine the time it takes for velocity to become zero at the maximum height. Understanding these changes in velocity helps in planning projectile timelines effectively.
The Role of Gravitational Acceleration
Gravitational acceleration is a constant that influences projectile motion significantly. On Earth, it is approximately \(9.8 \, \text{m/s}^2\), but for simplicity, this exercise uses \(10 \, \text{m/s}^2\). This is the rate at which an object accelerates downwards due to gravity.

Using gravitational acceleration in our calculations allows us to predict how motion progresses vertically:
  • It helps determine how fast the vertical velocity decreases as the object ascends, leading to a temporary stop at the peak height.
  • It is key in calculating the time interval of upward motion using the formula \[t = \frac{u}{g}\].
Gravitational acceleration is fundamental in determining motion characteristics and ensuring precise timing for sequential throws in this exercise.

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Most popular questions from this chapter

For motion of an object along the \(x\)-axis, the velocity \(v\) depends on the displacement \(x\) as \(v=3 x^{2}-2 x\), then what is the acceleration at \(x=2 \mathrm{~m}\). (1) \(48 \mathrm{~m} \mathrm{~s}^{-2}\) (2) \(80 \mathrm{~ms}^{-2}\) (3) \(18 \mathrm{~ms}^{-2}\) (4) \(10 \mathrm{~ms}^{-2}\)

A man swimming downstream overcomes a float at a point M. After travelling distance \(D\), he turned back and passed the float at a distance of \(D / 2\) from the point \(M\). Then the ratio of speed of swimmer with respect to still water to the speed of the river will be (1) 1 (2) 2 (3) 4 (4) 3

A train is moving at a constant speed \(V\) when its driver observes another train in front of him on the same track and moving in the same direction with constant speed \(v\). If the distance between the trains is \(x\), then what should be the minimum retardation of the train so as to avoid collision? (1) \(\frac{(V+v)^{2}}{x}\) (2) \(\frac{(V-v)^{2}}{x}\) (3) \(\frac{(V+v)^{2}}{2 x}\) (4) \(\frac{(V-v)^{2}}{2 x}\)

A particle is moving along the \(x\)-axis whose instantaneous speed is given by \(v^{2}=108-9 x^{2}\). The acceleration of the particle is (1) \(-9 x \mathrm{~ms}^{-2}\) (2) \(-18 x \mathrm{~ms}^{-2}\) (3) \(\frac{-9 x}{2} \mathrm{~ms}^{-2}\) (4) None of these

The average velocity of a body moving with uniform acceleration after travelling a distance of \(3.06 \mathrm{~m}\) is \(0.34\) \(\mathrm{ms}^{-1} .\) If the change in velocity of the body is \(0.18 \mathrm{~m} \mathrm{~s}^{-1}\) during this time, its uniform acceleration is (1) \(0.01 \mathrm{~ms}^{-2}\) (2) \(0.02 \mathrm{~m} \mathrm{~s}^{-2}\) (3) \(0.03 \mathrm{~m} \mathrm{~s}^{-2}\) (4) \(0.04 \mathrm{~m} \mathrm{~s}^{-2}\)
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