Suppose a coin is dropped from the top of the Empire State building in New York, which is 1,454 feet tall. The position function for free-falling objects is:
s(t) = −16t^2 + v_0t + s_0, where v_0 is the initial velocity and s_0 is the initial position.
1. Find the instantaneous velocities when t =1 and t = 3 (I've already solved)
2. What is the name of the theorem that says there must be at least one solution to Part 1?
3. Find the velocity of the coin just before it hits the ground.
3.
v(t) = -32t
time to reach ground:
-16t^2 + 1454 = 0
t^2 = 90.875
t = appr 9.53 seconds
v(9.53) = -32(9.53) ft/s = appr -305 ft/s
To find the velocity of the coin just before it hits the ground, we need to find the time at which the coin reaches the ground first. Since the initial position of the coin is at the top of the Empire State building, we can set s(t) equal to the height of the building, which is 1,454 feet.
So, we have the equation:
-16t^2 + v_0t + s_0 = 1,454
Since the coin is dropped from rest (v_0 = 0), we can simplify the equation to:
-16t^2 + 1,454 = 0
Solving this quadratic equation, we find two possible values for t, one of which will be the time at which the coin reaches the ground. However, since it is a physical scenario, we discard the negative value of t because it doesn't make sense in this context.
So, we are left with:
t = √(1,454/16) ≈ 9.02 seconds
Now, to find the velocity of the coin just before it hits the ground, we can substitute t = 9.02 seconds into the velocity function:
v(t) = -16t + v_0
Since the coin was initially dropped with no initial velocity (v_0 = 0), the equation simplifies to:
v(t) = -16t
Substituting the value of t = 9.02 seconds, we can find the velocity:
v(9.02) = -16(9.02) ≈ -144.32 feet per second
Therefore, the velocity of the coin just before it hits the ground is approximately -144.32 feet per second. Since the velocity is negative, it indicates that the coin is falling downward.
To find the velocity of the coin just before it hits the ground, we will use the position function and find the time at which the position is equal to zero.
The position function for free-falling objects is given by:
s(t) = −16t^2 + v_0t + s_0
To find the time at which the coin hits the ground, we set s(t) = 0 and solve for t:
0 = −16t^2 + v_0t + s_0
Since the coin is dropped from rest, the initial velocity v_0 is 0 and the equation simplifies to:
0 = −16t^2 + s_0
Now, substitute s_0 with the initial position of the coin, which is the height of the Empire State Building, 1454 feet:
0 = −16t^2 + 1454
Next, we can solve this quadratic equation to find the time at which the coin hits the ground. We can factor out a common factor of 2 and rewrite the equation as:
0 = 2(-8t^2 + 727)
Setting each factor equal to zero, we have:
-8t^2 + 727 = 0
Rearranging the terms, we get:
8t^2 = 727
Dividing both sides by 8, we have:
t^2 = 90.875
Taking the square root of both sides, we find:
t ≈ ±9.536
Since time cannot be negative in this context, we consider the positive value:
t ≈ 9.536 seconds
Now, we know that the velocity of the coin just before it hits the ground is the derivative of the position function with respect to time at t = 9.536 seconds. We can use the derivative to find the velocity.
Taking the derivative of the position function s(t), we get:
v(t) = -32t + v_0
Since the initial velocity v_0 is 0, the equation simplifies to:
v(t) = -32t
Substituting t = 9.536 seconds, we can find the velocity just before the coin hits the ground:
v(9.536) = -32(9.536)
Calculating the value gives us:
v(9.536) ≈ -304.352 feet per second
So, the velocity of the coin just before it hits the ground is approximately -304.352 feet per second.