The height of a projectile fired upward is given by the formula s = v0t – 16t2, where s is the height, v0 is the initial velocity and t is the time. Find the times at which a projectile with an initial velocity of 128 ft/s will be 144 ft above the ground. Round to the nearest hundredth of a second.
t = sec (smaller value)
t = sec (larger value)
So you are solving:
144 = 128t - 16t^2
16t^2 - 128t + 144 = 0
t^2 - 8t + 9 = 0
I will complete the square, (in this case faster than the formula)
t^2 - 8t + 16 = -9+16
(t-4)^2 = 7
t-4 = ±√7
t = 4 ± √7
I get appr 6.65 and 1.35
h = hi + Vi t - (1/2) g t^2
in ancient English units this is
h = Hi + Vi t - (1/2) 32 t^2
If
Hi = 0
Vi = 128 ft/s
then
h = 128 t - 16 t^2
for h = 144
144 = 128 t - 16 t^2
16 t^2 - 128 t + 144 = 0
t^2 - 8 t + 9 = 0
t = [ 8 +/- sqrt(64 -36) ]/2
t = 4 +/- sqrt 7
t = 1.35
t = 6.65
To find the times at which the projectile will be 144 ft above the ground, we need to solve the equation:
s = v0t - 16t^2
Substituting the given values:
s = 144 ft
v0 = 128 ft/s
144 = 128t - 16t^2
We rearrange the equation to get a quadratic equation:
16t^2 - 128t + 144 = 0
Now, we can solve this quadratic equation for t. We can either factorize it or use the quadratic formula:
Using the quadratic formula (where a = 16, b = -128, and c = 144):
t = (-b ± √(b^2 - 4ac)) / 2a
t = (-(-128) ± √((-128)^2 - 4(16)(144))) / (2(16))
Simplifying:
t = (128 ± √(16384 - 9216)) / 32
t = (128 ± √7168) / 32
t = (128 ± 84.77) / 32
Now, we calculate the two values of t separately:
For t = (128 + 84.77) / 32:
t = 212.77 / 32
t ≈ 6.65 seconds (larger value)
For t = (128 - 84.77) / 32:
t = 43.23 / 32
t ≈ 1.35 seconds (smaller value)
Therefore, the times at which the projectile with an initial velocity of 128 ft/s will be 144 ft above the ground are approximately:
t ≈ 1.35 seconds (smaller value)
t ≈ 6.65 seconds (larger value)
To find the times at which a projectile with an initial velocity of 128 ft/s will be 144 ft above the ground, we need to solve the equation s = v0t - 16t^2 for t, where s = 144 ft and v0 = 128 ft/s.
Substituting the given values into the equation, we have:
144 = 128t - 16t^2
Now we need to rearrange the equation into a quadratic form:
16t^2 - 128t + 144 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 16, b = -128, and c = 144. Substituting these values into the formula:
t = (128 ± √((-128)^2 - 4 * 16 * 144)) / (2 * 16)
Simplifying further:
t = (128 ± √(16384 - 9216)) / 32
t = (128 ± √7168) / 32
Now, we can calculate the values inside the square root:
√7168 ≈ 84.8528137424
To find the two possible values of t, we substitute the positive and negative square root values:
t = (128 + 84.8528137424) / 32 ≈ 6.6529 seconds (larger value)
t = (128 - 84.8528137424) / 32 ≈ 1.5965 seconds (smaller value)
Rounded to the nearest hundredth of a second, the times at which the projectile is 144 ft above the ground are:
t ≈ 6.65 seconds (larger value)
t ≈ 1.60 seconds (smaller value)