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.
Similar Questions
- 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).
- A ball is thrown vertically upward with a speed of 30 m/s. The magnitude of its displacement after 4 s will be ______ (Take \(g = 10 \text{ m/s}^2\).)
- 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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- A car, starting from rest, is moving with a constant acceleration \(3\text{ m/s}^2\). Find the distance travelled by this car in \(10\text{ s}\).
- A ball is thrown vertically upwards with initial velocity \(V_0\) and returns to its starting point in 6 seconds then the initial velocity with which the ball was thrown will be
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- A ball is thrown vertically upward with a speed of 30 m/s. The magnitude of its displacement after 4 s will be _____ (Take \(g = 10\ m/s^2\).)
A particle experiences constant acceleration for 20 s after starting from rest. If it travels a distance X 1, in the first 10 s and distance X 2in the remaining 10 s, then which of the following is true?
Important Questions from Kinematic equations for uniformly accelerated motion
Which of the following is an equation of motion?
I. u = v + at
II. 2as = v 2– u 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
- 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).
Which of the following is an equation of motion?
I. u = v + at
II. 2as = v 2– u 2
A stone is dropped from a ballon going up with a uniform velocity of 5m/sec. If the ballon was 50 m high, then the stone was dropped, the height the ballon from ground when stone hits the ground will be:
(g = 10 m/s 2)
