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.

Question

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)

Answer 1

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

Answer 2

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

Answer 3

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)

Answer 4

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)