A ball is thrown straight down from the top of a 220-foot building with an initial velocity of -22 feet per second. What is its velocity after 3 seconds? What is its velocity after falling 108 feet?
s = -16t^2 - 22t + 220
v = ds/dt = -32t - 22
so at t=3
v = -32(3) - 22 = -118 ft/sec
when s = 108
108 = -16t^2 - 22t + 220
16t^2 + 22t - 112 = 0
t = (-22 ± √7652)/32 = 2.046
so when t = 2.046
v = -32(2.046) - 22 = -87.472 ft/s
The process is correct but the t=2.046 should be substituted to the first derivative which is velocity function.
s = -16t^2 - 22t + 220
v = ds/dt = -32t - 22
so at t=3
v = -32(3) - 22 = -118 ft/sec
when s = 108
108 = -16t^2 - 22t + 220
0 = -16t^2 - 22t + 112
factor out a -2
0 = -2(8t^2 - 11t + 56)
use quadratic equation
(-11 ± √-1671)/16 = imaginary
use -2 to find v(t)
v = -32(2) - 22 = -86 ft/s
since it's asking for the velocity after it falls 108 feet, it isn't asking for when the position function is equal to 108, it wants it when it is 112 since it fell 108 of the 220 total feet leaving it at a y-level of 112. so you solve for "t" when the position function is equal to 112 and you get t=2. then you plug it into the velocity function to get -86 ft/s
where did t=3 come from
To find the velocity of the ball after a certain amount of time or distance, we need to use the equations of motion. The equation for velocity as a function of time is:
v = u + at
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
In this case, the ball is thrown straight down, so the acceleration due to gravity (a) is approximately -32 feet per second squared. The negative sign indicates that the acceleration is directed downward.
Let's calculate the velocity after 3 seconds:
Given:
u = -22 feet per second
a = -32 feet per second squared
t = 3 seconds
Substituting these values into the equation, we have:
v = -22 + (-32) * 3
v = -22 - 96
v = -118 feet per second
Therefore, the velocity of the ball after 3 seconds is -118 feet per second.
Now, let's find the velocity after falling 108 feet. To do this, we'll use the equation for displacement as a function of time:
s = ut + (1/2)at^2
where:
s = displacement
u = initial velocity
a = acceleration
t = time
In this case, we need to rearrange the equation to solve for time:
s = ut + (1/2)at^2
0 = u(t) + (1/2)at^2 - s
We can set up an equation using this rearranged equation to find the time it takes to fall 108 feet:
0 = -22(t) + (1/2)(-32)(t^2) - 108
To solve this quadratic equation, we can either factor it or use the quadratic formula. Factoring it may be a bit challenging, so let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = -16, b = -22, and c = -108. Substituting these values into the quadratic formula, we get:
t = (-(-22) ± √((-22)^2 - 4(-16)(-108))) / (2(-16))
Simplifying further, we have:
t = (22 ± √(484 - 6912)) / (-32)
Taking the square root and simplifying, we get:
t = (22 ± √(-6428)) / -32
Here, we encounter a problem—the term inside the square root (√(-6428)) is not a real number because we can't take the square root of a negative number. This means that the ball does not fall 108 feet.
Therefore, we cannot determine the velocity after falling 108 feet in this scenario.
s(t) = -16t^2 - 22t + 220
s'(t) = ds/dt = -32t - 22
so at t=3 sec.
s'(3) = -32(3) - 22 = -118 ft/sec
when s(t) = 108 ft.
108 = -16t^2 - 22t + 220
16t^2 + 22t - 112 = 0
t = (22 + √7652)/-32 = -3.421115966 = 3.421 sec.
so at 108ft, the time is 3.421 sec.
use 3.421 sec. to find s'(t) at 108 ft.
s'(3.421) = -32(3.421) - 22 = -131.472 ft/sec (<- velocity at the height of 108ft)
equation:
s = -16t^2 - 22t + 220
v = ds/dt = -32t - 22
so at t=3
v = -16(3) - 22 = -70 ft/sec
when s = 108
108 = -16t^2 - 22t + 220
16t^2 + 22t - 112 = 0
t = (-22 ± √7652)/32 = 2.046 sec or a negative time
so when t = 2.046
v = -16(2.046) - 22 = - 54 ft/s