A bolt comes loose from underneath an elevator that is moving upward at a speed of 7.6 m/s. The bolt reaches the bottom of the elevator shaft in 3.8 s.
1) How high up was the elevator when the bolt came loose?
2) What is the speed of the bolt when it hits the bottom of the shaft?
you have posted three of these motion problems without any demonstrated effort. We will be happy to critique your efforts.
To answer these questions, we can use the kinematic equations to analyze the motion of the bolt and the elevator.
1) To find the height the elevator was at when the bolt came loose, we need to calculate the initial height (h0). We can use the equation:
h = h0 + v0t + (1/2)at^2
Where:
h0 is the initial height
v0 is the initial velocity (which is zero when the bolt comes loose)
t is the time (3.8 s in this case)
a is the acceleration (which is constant and equal to the acceleration due to gravity, 9.8 m/s^2)
Since the bolt comes loose from underneath the elevator, we can assume it starts at the bottom of the elevator shaft, meaning h0 = 0. Therefore, the equation simplifies to:
h = (1/2)at^2
Plugging in the values we have:
h = (1/2) * 9.8 m/s^2 * (3.8 s)^2
Calculating this will give you the height the elevator was at when the bolt came loose.
2) To find the speed of the bolt when it hits the bottom of the shaft, we can use the equation:
v = v0 + at
Where:
v is the final velocity of the bolt (which is what we're trying to find)
v0 is the initial velocity of the bolt (which is zero when the bolt comes loose)
t is the time it takes for the bolt to reach the bottom of the shaft (3.8 s in this case)
a is the acceleration due to gravity (-9.8 m/s^2 in this case, taking downward as the negative direction)
Plugging in the values we have:
v = 0 + (-9.8 m/s^2) * (3.8 s)
Calculating this will give you the speed of the bolt when it hits the bottom of the shaft.