A piece of gum comes lose from underneath an elevator that is moving upward at a speed of 6.30 m/s. The gum reaches the bottom of the elevator shaft in 3.00 s. With what speed does the gum hit the bottom of the shaft?
V = Vo + g*t.
Vo = - 6.3 m/s.
g = +9.8 m/s.
t = 3 s.
V = ?
To determine the speed at which the gum hits the bottom of the elevator shaft, we can use the equations of linear motion.
Let's denote the initial velocity of the gum as 'u', the final velocity as 'v', the time taken as 't', and the acceleration as 'a'.
Given:
The elevator is moving upward at a speed of 6.30 m/s, which means the initial velocity of the gum relative to the ground is also 6.30 m/s.
The time taken for the gum to reach the bottom of the shaft is 3.00 s.
To find the speed at which the gum hits the bottom of the shaft, we need to calculate the final velocity 'v'.
We can use the equation of motion:
v = u + at
Since the gum is moving vertically downward, the acceleration due to gravity, denoted as 'g', will act in the downward direction. The value of 'g' is approximately 9.8 m/s^2.
Now, let's substitute the known values into the equation:
v = 6.30 m/s + (9.8 m/s^2)(3.00 s)
v = 6.30 m/s + 29.4 m/s
v = 35.7 m/s
Therefore, the speed at which the gum hits the bottom of the shaft is approximately 35.7 m/s.