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Chapter 4: Problem 41

Drops of water fall at regular intervals from roof of a building of height \(H=16 \mathrm{~m}\), the first drop striking the ground at the same moment as the fifth drop detaches itself from the roof. The distances between separate drops in air as the first drop reaches the ground are (1) \(1 \mathrm{~m}, 5 \mathrm{~m}, 7 \mathrm{~m}, 3 \mathrm{~m}\) (2) \(1 \mathrm{~m}, 3 \mathrm{~m}, 5 \mathrm{~m}, 7 \mathrm{~m}\) (3) \(1 \mathrm{~m}, 3 \mathrm{~m}, 7 \mathrm{~m}, 5 \mathrm{~m}\) (4) None of the above

Short Answer

Expert verified
The distances between drops are 1 m, 3 m, 5 m, and 7 m.

Step by step solution

01

Calculate Time for First Drop to Hit Ground

We start by using the free fall formula for the first drop, which strikes the ground when the fifth drop releases. Under constant acceleration due to gravity ( g = 9.8 \ \mathrm{m/s}^2 ), the formula to calculate the time t it takes for the first drop to hit the ground is: \[ H = \frac{1}{2} g t^2 \]Given \ H = 16 \ \mathrm{m} , substitute the values: \[ 16 = \frac{1}{2} \times 9.8 \times t^2 \]Solving for t yields t = \sqrt{\frac{32}{9.8}} \approx 1.8 \ \! \mathrm{seconds} \ along with the necessary calculations.
02

Calculate Interval Between Drops

Since the fifth drop is released just as the first drop strikes the ground, the time interval between the release of consecutive drops is the time taken for the first to hit the ground divided by 4 (as there are four intervals before the fifth drop): \[ \Delta t = \frac{1.8}{4} = 0.45 \!\mathrm{seconds} \]. Thus, each drop is released every 0.45 \ \mathrm{seconds} .
03

Calculate Distances of Drops in Air

We determine the distance fallen by each drop when the first hits the ground. For each drop after the first, use the formula for distance fallen in time t_k = k \cdot \Delta t seconds:1. Second Drop: t_1 = 0.45 seconds. \[ d_2 = \frac{1}{2} \times 9.8 \times (0.45)^2 \approx 1 \ \mathrm{m} \]2. Third Drop: t_2 = 0.9 seconds. \[ d_3 = \frac{1}{2} \times 9.8 \times (0.9)^2 \approx 4 \ \mathrm{m} \]3. Fourth Drop: t_3 = 1.35 seconds. \[ d_4 = \frac{1}{2} \times 9.8 \times (1.35)^2 \approx 9 \ \mathrm{m} \]4. Fifth Drop: t_4 = 1.8 seconds, but only released before the impact. \[ d_5 = 0 \ \mathrm{m} \].
04

Calculate Distances Between Drops

The distances between each pair of consecutive drops are achieved by subtracting adjacent falls:1. Between Second and Third: \[ 4 - 1 = 3 \ \mathrm{m} \]2. Between Third and Fourth: \[ 9 - 4 = 5 \ \mathrm{m} \]3. Between Fourth and Fifth (at initial): \[ 16 - 9 = 7 \ \mathrm{m} \]Thus, the distances between drops are 1 \mathrm{~m}, 3 \mathrm{~m}, 5 \mathrm{~m}, 7 \mathrm{~m}.

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Key Concepts

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

Free Fall
Understanding free fall is key to comprehending how objects behave when dropped from a height, such as droplets of water falling from a building roof. In free fall, gravity is the only force acting on the object. This means that other forces, like air resistance, are negligible. The object accelerates downward at a constant rate, which on Earth is approximately 9.8 meters per second squared (m/s虏). During free fall, the motion can be described by the equation of motion:
  • \[ H = \frac{1}{2}gt^2 \] where \(H\) stands for the height, \(g\) is the acceleration due to gravity, and \(t\) represents time.
Remember, this describes a situation where an object starts from rest. Thus, if you know the height from which an object like a water droplet falls, you can calculate the time it takes to hit the ground.
Vertical Motion
Vertical motion refers specifically to the motion of objects moving up or down without any horizontal movement. In our exercise with water drops, as each drop falls, it becomes an illustration of vertical motion in real-time. These droplets have an initial velocity of zero because they start from rest on the roof. As time progresses, the velocity of each drop increases due to gravitational pull. The motion of each drop is predictable using physics equations for uniformly accelerated motion. If you were to measure how fast a drop is moving when it hits the ground, you could use the equation:
  • \[ v = gt \]
Where \(v\) is the velocity of the drop when it reaches the ground. It's fascinating how predictable such movements can be with the right calculations!
Time Interval Calculations
Time interval calculations in kinematics often involve determining how long it takes for an event to occur, and how those timings relate to sequences of events. For instance, when calculating the release and fall times of our water drops, we divided the total fall time of the first drop by the number of intervals between successive drops. By doing so, we were able to find the precise time at which each drop was released.Understanding time intervals helps in predicting and analyzing sequences in motion. This concept applied to our exercise clarifies why the fifth drop is just released as the first one hits the ground. Each interval is consistent at:
  • \[ \Delta t = \frac{1.8}{4} = 0.45 \ \! \mathrm{seconds} \]
This detail allows for a uniform distribution of droplet release from the roof, resulting in the fascinating arrangements of droplets in mid-air.

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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 particle is moving along the \(x\)-axis whose acceleration is given by \(a=3 x-4\), where \(x\) is the location of the particle. At \(t=0\), the particle is at rest at \(x=4 / 3 \mathrm{~m} .\) The distance travelled by the particle in \(5 \mathrm{~s}\) is (1) zero (2) \(42 \mathrm{~m}\) (3) Infinite (4) None of these

The body will speed up if (1) Velocity and acceleration are in the same direction. (2) Velocity and acceleration are in opposite directions. (3) Velocity and acceleration are in perpendicular direction. (4) Velocity and acceleration are acting at acute angle w.r.t. each other.

A drunkard is walking along a straight road. He takes five steps forward and three steps backward and so on. Each step is \(1 \mathrm{~m}\) long and takes \(1 \mathrm{~s}\). There is a pit on the road \(11 \mathrm{~m}\) away from the starting point. The drunkard will fall into the pit after (1) \(29 \mathrm{~s}\) (2) \(21 \mathrm{~s}\) (3) \(37 \mathrm{~s}\) (4) \(31 \mathrm{~s}\)

For a particle moving along the \(x\)-axis, mark the correct statements (1) If x is positive and is increasing with the time, then average velocity of the particle is positive. (2) If x is negative and becoming positive after some time then the velocity of the particle is always positive. (3) If x is negative and becoming less negative as time passes then the average velocity of the particle is positive (4) If x is positive and is increasing with time, then the velocity of the particle is always positive.
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