An elevator is travelling upwards at 15m/s2 and the hoisting cable is cut when it is 40 m from the ground. Determine the maximum height reached by the elevator & the velocity just before it hits the ground.
To solve this problem, we need to use the equations of motion under constant acceleration.
First, let's determine the time it takes for the elevator to reach the maximum height. We can use the equation:
\(v^2 = u^2 + 2as\),
where:
- \(v\) is the final velocity (which is 0 m/s at the maximum height),
- \(u\) is the initial velocity of the elevator (15 m/s),
- \(a\) is the acceleration (which the problem states is 15 m/s^2),
- \(s\) is the displacement (40 m).
Substituting the known values into the equation, we have:
\(0^2 = 15^2 + 2 \cdot 15 \cdot s\),
\(0 = 225 + 30s\),
\(30s = -225\),
\(s = -7.5 \, \text{m}\).
Since the displacement (\(s\)) is negative, it means the elevator is moving upwards. Therefore, the elevator will reach its maximum height at \(7.5 \, \text{m}\) above its initial position.
Next, let's determine the maximum height reached by the elevator. We can use the equation:
\(s = ut + \frac{1}{2}at^2\),
where:
- \(s\) is the maximum height,
- \(u\) is the initial velocity (15 m/s),
- \(a\) is the acceleration (15 m/s^2),
- \(t\) is the time taken to reach the maximum height.
Since we have already found \(s = 7.5 \, \text{m}\), we need to solve for \(t\). Rearranging the equation, we have:
\(\frac{1}{2}at^2 = s - ut\),
\(7.5 = 15t - \frac{1}{2} \cdot 15 \cdot t^2\),
\(7.5 = 15t - 7.5t^2\).
Rearranging the equation, we get a quadratic equation:
\(7.5t^2 - 15t + 7.5 = 0\).
Simplifying the equation, we get:
\(t^2 - 2t + 1 = 0\),
\((t - 1)^2 = 0\).
The only solution is \(t = 1\).
Therefore, the elevator takes 1 second to reach the maximum height.
To find the velocity just before it hits the ground, we can use the equation:
\(v = u + at\),
where:
- \(v\) is the final velocity just before hitting the ground,
- \(u\) is the initial velocity (15 m/s),
- \(a\) is the acceleration (15 m/s^2),
- \(t\) is the total time taken (which equals twice the time taken to reach the maximum height, so \(2t\)).
Substituting the values, we have:
\(v = 15 + 15 \cdot 2 \cdot 1\),
\(v = 15 + 30\),
\(v = 45 \, \text{m/s}\).
Therefore, the velocity just before the elevator hits the ground is \(45 \, \text{m/s}\).