1. The problem statement, all variables and given/known data
a ball is thrown straight down from the top of a 220 feet building with an initital velocity of -22ft/s. What is the velocity of the ball after 3 seconds? What is the velocity of the ball after falling 108 feet?
2. Relevant equations
s(t) = -16t^2 + Vi(t) + si vi = initial velocity, si = initial position
3. The attempt at a solution
s(t) = -16t^2 +Vi(t) +si
= -16t^2 - 22ft/s + 200ft
V(t) = -32(t) - 22ft/s
V(3) = -96 ft/s - 22ft/s = -118 ft/s
I can seem to figure out how to determine the velocity of the ball after it has dropped 108ft.
I should be able to do this, but i'm missing something, somewhere on what should be pretty basic.
Thanks for any guidance!
done under your previous post as Audrey
http://www.jiskha.com/display.cgi?id=1328273218
To determine the velocity of the ball after it has dropped 108 feet, we need to consider the equation for displacement:
s(t) = -16t^2 + Vi(t) + si
Here, s(t) represents the displacement at time t, Vi represents the initial velocity, and si represents the initial position.
Since we want to find the time when the ball has dropped 108 feet, we can set s(t) equal to -108 and solve for t.
-108 = -16t^2 - 22t + 220
Simplifying the equation, we get:
16t^2 + 22t - 328 = 0
Now, we can solve this quadratic equation to find the value of t.
Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = 16, b = 22, and c = -328.
t = (-22 ± √(22^2 - 4(16)(-328))) / (2(16))
Simplifying further:
t = (-22 ± √(484 + 20992)) / 32
t = (-22 ± √(21476)) / 32
t ≈ -1.482 or t ≈ 6.482
Since time cannot be negative in this context, we can ignore the negative value.
Therefore, after the ball has fallen 108 feet, the time is approximately 6.482 seconds.
To find the velocity of the ball after falling 108 feet, we can substitute this value of t into the equation for velocity:
V(t) = -32(t) - 22 ft/s
V(6.482) = -32(6.482) - 22 ft/s
V(6.482) ≈ -212.544 - 22 ft/s
V(6.482) ≈ -234.544 ft/s
Hence, the velocity of the ball after falling 108 feet is approximately -234.544 ft/s.
To determine the velocity of the ball after it has fallen 108 feet, we can use the equation for displacement:
s(t) = -16t^2 + Vi(t) + si
In this case, the initial position (si) is 220 feet (the height of the building) and the initial velocity (Vi) is -22 ft/s (since the ball is thrown straight down). We also know that the displacement (s(t)) is -108 feet, as the ball has fallen 108 feet.
So, we can plug these values into the equation:
-108 = -16t^2 - 22t + 220
Now, we can solve this quadratic equation for t. Rearranging the equation, we have:
16t^2 + 22t - 328 = 0
To solve this quadratic equation, you can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 16, b = 22, and c = -328. Plugging in these values, we have:
t = (-22 ± √(22^2 - 4*16*(-328))) / 2*16
Now, we can calculate the values inside the square root:
t = (-22 ± √(484 + 20992)) / 32
t = (-22 ± √21476) / 32
We have two possible solutions for t, but since we're dealing with time, we only consider the positive solution:
t = (-22 + √21476) / 32 ≈ 2.55 seconds
Now that we have the time, we can substitute it back into the equation for velocity:
V(t) = -32t - 22
V(2.55) ≈ -32(2.55) - 22 ≈ -81.6 ft/s
Therefore, the velocity of the ball after falling 108 feet is approximately -81.6 ft/s.