A stone is dropped from a tower 100 meters above the ground. The stone falls past ground level and into a well. It hits the water at the bottom of the well 5.00 seconds after being dropped from the tower. Calculate the depth of the well. Given: g = -9.81 meters/second2.
h= h₁ - h₂= gt²/2- 100=9.8•5²/2 -100=122.5 – 100 = 22.5m
timetodrop=distance/avgspeed
avg speed=1/2 vf=1/2 at
timeto derop=(100m+d)/(1/2 9.8 *t)
5*(4.9*5)=(100+d)
d=4.9*25 -100
22.5
distance= 1/2*gravity*time^2-total distance
d=1/2*9.8*5^2-100
d=22.5
To calculate the depth of the well, we need to find the time it takes for the stone to reach the ground level.
First, let's use the kinematic equation for vertical motion:
Δy = v0*t + (1/2)g*t^2
Where:
Δy = displacement (depth of the well)
v0 = initial velocity (0 m/s because the stone is dropped)
t = time in seconds
g = acceleration due to gravity (-9.81 m/s^2)
Since the stone is dropped from rest, v0 = 0, and the equation simplifies to:
Δy = (1/2)g*t^2
We can rearrange the equation to solve for t:
t = sqrt((2*Δy)/g)
Given that Δy = 100 meters, we substitute it into the equation:
t = sqrt((2*100)/(-9.81))
Calculating:
t = sqrt(-204.08)
Here, we encounter a problem because we cannot take the square root of a negative number when dealing with real values. Hence, the stone does not reach the ground level within 5.00 seconds. Therefore, it must hit the water at the bottom of the well before reaching the ground level.
Since the depth of the well is less than 100 meters, we cannot determine the exact depth based on the information given.