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!
Your equation
s(t) = 16t^2 - 22t + 220 is correct and tells you
how high the object is at any time t
But they are asking for the velocity after it has fallen 108 feet, not when it is 108 feet high
It started at 220, so after falling 108 feet, the height will be 112
so 112 = -16t^2 - 22t + 220
16t^2 + 22t - 108 = 0
t = (-22 ± √7396)/32
= 2 or a negative, (ahhh, it would have factored)
so take v(2) , that should give you the answer you need
check:
when t=0 , s(0) = 220 -- it has fallen 0 feet
when t=1 , s(1) = 182 -- it has fallen 38 ft
when t=2 , s(2) = 112 -- it has fallen 108 ft
To determine the velocity of the ball after it has fallen 108 feet, you can use the equation for position:
s(t) = -16t^2 + Vi(t) + si
Here, si is the initial position, which is the height of the building (220 ft) and Vi is the initial velocity (-22 ft/s).
To find the time it takes for the ball to fall 108 feet, we need to solve the equation:
-16t^2 - 22t + 220 = 108
Simplifying the equation, we have:
-16t^2 - 22t + 112 = 0
To solve this quadratic equation, you can either factor it or use the quadratic formula. In this case, factoring is a bit tricky, so we'll use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = -16, b = -22, and c = 112. Substituting these values into the quadratic formula:
t = (-(-22) ± √((-22)^2 - 4(-16)(112))) / (2(-16))
Simplifying further:
t = (22 ± √(484 + 7168)) / (-32)
t = (22 ± √7652) / -32
Now, calculate the two possibilities:
t = (22 + √7652) / -32 and t = (22 - √7652) / -32
The correct value for t will be the positive one because the ball is falling downwards and time cannot be negative. Therefore, t = (22 + √7652) / -32.
Now, we can substitute this value of t into the velocity equation to find the velocity after falling 108 feet:
V(t) = -32(t) - 22
V(22 + √7652) / -32 = -32((22 + √7652) / -32) - 22
Simplifying:
V(22 + √7652) / -32 = -22 + √7652
So, the velocity of the ball after falling 108 feet is approximately -22 + √7652 ft/s.
To determine the velocity of the ball after falling 108 feet, we need to find the time it takes for the ball to reach that position. We can use the equation for the position of the ball, s(t) = -16t^2 - 22t + 200, where s(t) is the position of the ball at time t.
To find the time it takes for the ball to reach 108 feet, we set s(t) equal to 108 and solve for t:
-16t^2 - 22t + 200 = 108
Simplifying the equation, we have:
-16t^2 - 22t + 92 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = -16, b = -22, and c = 92. Plugging in these values, we can calculate t:
t = (-(-22) ± √((-22)^2 - 4(-16)(92))) / 2(-16)
= (22 ± √(484 + 5888)) / (-32)
= (22 ± √(6372)) / (-32)
Now we have two possible solutions for t, one of which will be the time it takes for the ball to reach 108 feet. We can calculate both possibilities:
t1 = (22 + √(6372)) / (-32)
t2 = (22 - √(6372)) / (-32)
Calculating these values, we find approximately:
t1 ≈ -0.52 seconds
t2 ≈ 3.40 seconds
Since time cannot be negative in this context, we discard t1 ≈ -0.52 seconds. Therefore, the time it takes for the ball to reach 108 feet is approximately t ≈ 3.40 seconds.
Finally, we can use this value of t in the velocity equation, V(t) = -32t - 22ft/s, to find the velocity of the ball at 3.40 seconds:
V(3.40) = -32(3.40) - 22ft/s
= -108.8ft/s - 22ft/s
≈ -130.8ft/s
Therefore, the velocity of the ball after falling 108 feet is approximately -130.8 ft/s.