An elevator with a broken cable is in free fall. If it was initially moving upward at 5 m/s, how long does it take for the elevator to hit the ground? What is the velocity just before impact?
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To solve this problem, we need to consider the principles of motion in free fall.
We know that the elevator is initially moving upward at 5 m/s, which means its velocity is positive. However, when the cable is broken, the elevator starts to fall downward due to gravity. Gravity will cause the elevator to accelerate at a rate of 9.8 m/s².
We can use the following kinematic equation to find the time it takes for the elevator to hit the ground:
h = ut + (1/2)gt²
Where:
h = height or distance fallen
u = initial velocity
g = acceleration due to gravity
t = time
In this case, the initial height of the elevator is not given, but it is not required to solve the problem. We are interested in the time it takes for the elevator to hit the ground, which means h = 0 (as it has fallen to the ground).
0 = (5)t + (1/2)(9.8)t²
Now, let's solve this quadratic equation:
0 = 5t + 4.9t²
Rearranging the equation:
4.9t² + 5t = 0
Factoring out a t:
t (4.9t + 5) = 0
The equation will be true if t = 0 (which is the initial condition), or if 4.9t + 5 = 0.
Solving 4.9t + 5 = 0:
4.9t = -5
t = -5/4.9
t ≈ -1.02 seconds
The negative result indicates that the elevator has already fallen for 1.02 seconds, so we take the positive value for t:
t ≈ 1.02 seconds
Therefore, it takes approximately 1.02 seconds for the elevator to hit the ground.
To find the velocity just before impact, we can use the kinematic equation:
v = u + gt
Where:
v = final velocity
u = initial velocity
g = acceleration due to gravity
t = time
v = 5 + (9.8)(1.02)
v = 5 + 9.996
v ≈ 14.996 m/s
Therefore, the velocity just before impact is approximately 14.996 m/s (downward direction).
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