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Question

A ball, initially at rest, falls freely from the top of a building and reaches a maximum velocity of \(40\text{ m/s}\). Find the height of the building. (Use acceleration due to gravity, \(g = 10\text{ m/s}^2\))

This question was previously asked in
RRB ALP 2018 CBT 2 Fitter Question Paper (21-Jan-2019) (Shift 3)
The correct answer is
$80\text{ m}$

Free Fall Building Height Calculation

Problem Analysis

The problem involves a ball in free fall, starting from rest. We are given the final velocity reached and the acceleration due to gravity. We need to determine the height from which the ball fell, which corresponds to the height of the building.

  • Initial velocity (\(u\)): \(0\text{ m/s}\) (since the ball starts from rest)
  • Final velocity (\(v\)): \(40\text{ m/s}\) (maximum velocity reached)
  • Acceleration (\(a\)): \(g = 10\text{ m/s}^2\) (acceleration due to gravity)
  • Displacement (\(s\)): Height of the building (\(h\)), which needs to be calculated.

Kinematic Equation for Height

We can use a standard kinematic equation that relates initial velocity, final velocity, acceleration, and displacement. The appropriate equation is:

\(v^2 = u^2 + 2as\)

Calculation Steps

Substitute the known values into the kinematic equation:

\((40\text{ m/s})^2 = (0\text{ m/s})^2 + 2 \times (10\text{ m/s}^2) \times h\)

Simplify the equation:

\(1600\text{ m}^2/\text{s}^2 = 0 + (20\text{ m/s}^2) \times h\)

\(1600\text{ m}^2/\text{s}^2 = (20\text{ m/s}^2) \times h\)

Now, solve for the height (\(h\)):

\(h = \frac{1600\text{ m}^2/\text{s}^2}{20\text{ m/s}^2}\)

\(h = 80\text{ m}\)

Therefore, the height of the building is 80 meters.

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Similar Questions

  1. A ball thrown up vertically returns to the ground after 10 second. Find the velocity with which it was thrown up? (if g = 10 m/s2).
  2. A ball is thrown vertically upward with a speed of 30 m/s. The magnitude of its displacement after 4 s will be ______ (Take \(g = 10 \text{ m/s}^2\).)
  3. A body falling from rest has a velocity '\(v\)' after it falls through a distance '\(h\)'. The distance it has to fall down further, for its velocity to become double, is _____ times '\(h\)' .
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Important Questions from Kinematic equations for uniformly accelerated motion

  1. Which of the following is an equation of motion?

    I. u = v + at

    II. 2as = v 2– u 2

  2. A ball is thrown vertically upward with a speed of 40 m/s. The time taken by the ball to reach the maximum height would be approximately
  3. A ball thrown up vertically returns to the ground after 10 second. Find the velocity with which it was thrown up? (if g = 10 m/s2).
  4. If the distance travelled by a body in the $n^{th}$ second is given by $(7 + 5n)$ m, then find the initial velocity and acceleration of the body respectively.

  5. Which of the following is an equation of motion?

    I. u = v + at

    II. 2as = v 2– u 2

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