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Chapter 2: Problem 23

One body is dropped, while a second body is thrown downward with an initial velocity of \(1 \mathrm{~ms}^{-1}\) simultaneously. The separation between these is \(1.8 \mathrm{~m}\) after a time (A) \(4.5 \mathrm{~s}\) (B) \(9 \mathrm{~s}\) (C) \(1.8 \mathrm{~s}\) (D) \(36 \mathrm{~s}\)

Short Answer

Expert verified
The separation between the two bodies is 1.8 m when the time is \(1.8 \mathrm{~s}\). Therefore, the correct answer is (C) \(1.8 \mathrm{~s}\).

Step by step solution

01

Understanding the situation

For body 1, which is dropped, the initial velocity (u1) is zero, and the acceleration (a1) is equal to the acceleration due to gravity (g = 9.8 m/s虏, downward). For body 2, thrown downward with an initial velocity (u2) of 1 m/s, the acceleration (a2) is also equal to g.
02

Calculate the distances traveled by each body

We can use the equation of motion: \(s = ut + \frac{1}{2}at^2\), where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time. For body 1: \(s1 = u1t + \frac{1}{2}g t^2\) For body 2: \(s2 = u2t + \frac{1}{2}g t^2\) Now, we plug in the values for body 1 and body 2: Body 1: \(s1 = 0 + \frac{1}{2}(9.8)t^2\) Body 2: \(s2 = (1)t + \frac{1}{2}(9.8)t^2\)
03

Calculate the separation between the two bodies

The separation, d, between the two bodies is given by the difference in their distances: \(d = s2 - s1\) Substituting the expressions for s1 and s2, we get: \(d = [ (1)t + \frac{1}{2}(9.8)t^2\] - [0 + \frac{1}{2}(9.8)t^2\)
04

Simplify the expression for the separation

Simplifying the expression for the separation, we get: \(d = (1)t\)
05

Find the time that matches the given separation

We are given that the separation between the two bodies is 1.8 m. Therefore, we have: \(1.8 = (1)t\) Now, we can find the time by solving for t: \(t = 1.8 \mathrm{~s}\)
06

Choose the correct option

The correct option for the time when the separation between the two bodies is 1.8 m is: (C) \(1.8 \mathrm{~s}\)

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Most popular questions from this chapter

The horizontal range of a projectile fired at an angle of 15 is \(50 \mathrm{~m}\). If it is fired with the same speed at an angle of 45 , its range will be (A) \(60 \mathrm{~m}\) (B) \(71 \mathrm{~m}\) (C) \(100 \mathrm{~m}\) (D) \(141 \mathrm{~m}\)

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