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Chapter 3: Problem 57

Balls are thrown vertically upward in such a way that the next ball is thrown when the previous one is at the maximum height. If the maximum height is \(5 \mathrm{~m}\), the number of balls thrown per minute will be: (Take \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) ) (a) 60 (b) 40 (c) 50 (d) 120

Short Answer

Expert verified
30 throws per minute, but no given option matches.

Step by step solution

01

Understanding Maximum Height of a Projectile

The maximum height reached by a projectile thrown upward with initial velocity \(u\) is given by the equation \( h = \frac{u^2}{2g} \). In this exercise, \( h \) is given as 5 meters and \( g \) is 10 m/s虏. We need to solve for \( u \), the initial velocity.
02

Calculate Initial Velocity

Using the formula \( h = \frac{u^2}{2g} \), we substitute \( h = 5 \) m and \( g = 10 \) m/s虏 to solve for \( u \):\[5 = \frac{u^2}{2 \times 10}\]Solving for \( u \), we get:\[u^2 = 5 \times 20 = 100 \quad \Rightarrow \quad u = \sqrt{100} = 10 \text{ m/s}\]
03

Calculate Time to Reach Maximum Height

With the initial velocity \( u = 10 \) m/s, we use the formula for time \( t \) to reach the maximum height: \( t = \frac{u}{g} \).\[t = \frac{10}{10} = 1 \text{ second}\]
04

Calculate Sinlge Throw Time

Since each complete throw (up and down) takes twice the time to reach the maximum height, the time taken for one complete throw is:\[\text{Total time for one throw} = 2 \times 1 \text{ second} = 2 \text{ seconds}\]
05

Calculate Number of Throws per Minute

Since each throw takes 2 seconds, the number of throws per minute is calculated as:\[\text{Throws per minute} = \frac{60}{2} = 30 \]
06

Correct Option

The answer based on our calculations, 30 throws per minute, does not match any of the given options. Please review the calculation steps to ensure accuracy or reevaluate the provided options.

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Key Concepts

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

Maximum Height Formula
When dealing with projectile motion, especially when objects are thrown vertically upwards, understanding how to calculate the maximum height is crucial. This is done using the formula:
  • \( h = \frac{u^2}{2g} \)
where:
  • \( h \) is the maximum height.
  • \( u \) is the initial velocity of the projectile.
  • \( g \) is the acceleration due to gravity, which is usually around \( 9.81 \, \text{m/s}^2 \) but can be approximated to \( 10 \, \text{m/s}^2 \) for easier calculations.
In this exercise, you are given that the maximum height \( h \) is \( 5 \, \text{m} \) and \( g \) is \( 10 \, \text{m/s}^2 \), which allows us to find the initial velocity.
Understanding this formula is fundamental in solving various problems involving the motion of projectiles.
Initial Velocity Calculation
The initial velocity of a projectile plays an important role in determining its motion. To find the initial velocity \( u \), we rearrange the maximum height formula:
  • \( h = \frac{u^2}{2g} \)
to solve for \( u \):
  • \( u = \sqrt{2gh} \)
Given that \( h = 5 \, \text{m} \) and \( g = 10 \, \text{m/s}^2 \), plug these values into the formula:
  • \( u = \sqrt{2 \times 10 \times 5} = \sqrt{100} = 10 \, \text{m/s} \)
This calculation provides the speed at which the projectile must be thrown to reach a height of 5 meters. Understanding how to derive and plug into this formula is vital for problems involving projectiles.
Time of Flight
The time it takes for a projectile to reach its maximum height can be determined by using the formula:
  • \( t = \frac{u}{g} \)
where:
  • \( u \) is the initial velocity of the projectile.
  • \( g \) is the acceleration due to gravity.
With an initial velocity \( u = 10 \, \text{m/s} \) and gravity \( g = 10 \, \text{m/s}^2 \), the calculation becomes:
  • \( t = \frac{10}{10} = 1 \, \text{second} \)
This time only accounts for the ascent. The total time of flight, which includes the descent, is twice this value, giving us:
  • \( 2 \times t = 2 \, \text{seconds} \)
Understanding the time of flight helps in predicting projectile behavior throughout its motion.
Throw Frequency Calculation
Projectiles, like balls in the exercise, can be thrown at specific intervals to achieve a continuous pattern. To determine how many throws can be completed in a given time, we calculate the throw frequency. With each throw taking a total time of 2 seconds, we find the number of throws per minute using the formula:
  • \( \text{Throws per minute} = \frac{60}{\text{Time for one throw}} \)
Given the time for one throw is 2 seconds:
  • \( \text{Throws per minute} = \frac{60}{2} = 30 \)
While the provided options suggest otherwise, this calculation gives us a logical basis for understanding how frequent the throws can actually be. Knowing how to calculate throw frequency is useful in applications involving repeated projectile launches.

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Most popular questions from this chapter

A particle is projected with speed \(v\) at an angle \(\theta\) \(\left(0<\theta<\frac{\pi}{2}\right)\) above the horizontal from a height \(H\) above the ground. If \(v=\) speed with which particle hits the ground and \(t=\) time taken by particle to reach ground, then: (a) as \(\theta\) increases, \(v\) decreases and \(t\) increases (b) as \(\theta\) increases, \(v\) increases and \(t\) increases (c) as \(\theta\) increases, \(v\) remains same and \(t\) increases (d) as \(\theta\) increases, \(v\) remains same and \(t\) decreases

When acceleration of a particle is \(a=f(t)\), then: (a) the velocity, starting from rest is \(\int_{n}^{i} f(t) d t\) (b) velocity may be constant (c) the velocity must not be function of time (d) the speed may be constant with respect to time

A particle moving with a uniform acceleration along a straight line covers distances \(a\) and \(b\) in successive intervals of \(p\) and \(q\) second. The acceleration of the particle is : (a) \(\frac{p q(p+q)}{2(b p-a q)}\) (b) \(\frac{2(a q-b p)}{p q(p-q)}\) (c) \(\frac{b p-a q}{p q(p-q)}\) (d) \(\frac{2(b p-a q)}{p q(p+q)}\)

A body starts from rest and moves with a constant acceleration. The ratio of distance covered in the \(n\) th second to the distance covered in \(n\) second is : (a) \(\frac{2}{n}-\frac{1}{n^{2}}\) (b) \(\frac{1}{n^{2}}-\frac{1}{n}\) (c) \(\frac{2}{n^{2}}-\frac{1}{n}\) (d) \(\frac{2}{n}+\frac{1}{n^{2}}\)

An object moves, starting from rest through a resistive medium such that its acceleration is related to velocity as \(a=3-2 v\). Then : (a) the terminal velocity is \(1.5\) unit (b) the terminal velocity is 3 unit (c) the slope of \(a-v\) graph is not constant (d) initial acceleration is 2 unit
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