The height of a projectile fired upward is given by the formula s = v0t – 16t^2, 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 176 ft above the ground. Round to the nearest hundredth of a second.
so your equation would clearly be
s = 128t - 16t^2
you want that to be 176
176 = 128t - 16t^2
16t^2 - 128t + 176 = 0
all 3 terms divide by 16, how nice
t^2 - 8t + 11 = 0
Use your favourite method to solve this.
You will get 2 positive answers, explained by the fact that the height will be
176 ft on its way up (the smaller t value) and 176 ft on its way down (the larger t)
To find the times at which the projectile will be 176 ft above the ground, we need to set the height equation equal to 176 and solve for time (t).
The height equation is: s = v0t - 16t^2
Substitute s = 176 and v0 = 128 into the equation:
176 = 128t - 16t^2
Rearrange the equation to put it in standard quadratic form:
16t^2 - 128t + 176 = 0
To solve for t, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 16, b = -128, and c = 176. Plug these values into the formula:
t = (-(-128) ± √((-128)^2 - 4 * 16 * 176)) / (2 * 16)
Simplifying further:
t = (128 ± √(16384 - 11264)) / 32
t = (128 ± √(51120)) / 32
Now, let's calculate the values of t:
t1 = (128 + √(51120)) / 32
t2 = (128 - √(51120)) / 32
Round the values to the nearest hundredth of a second:
t1 ≈ 4.43 seconds
t2 ≈ 4.57 seconds
Therefore, a projectile with an initial velocity of 128 ft/s will be 176 ft above the ground at approximately 4.43 seconds and 4.57 seconds after being fired upward.