A classic physics problem states that if a projectile is shot vertically up into the air with an initial velocity of 127 feet per second from an initial height of 114 feet off the ground, then the height of the projectile, h, in feet, t seconds after its shot is given by the equation:
h=-16t^2+127t+114
Find the two points in time when the object is 151 feet above the ground. Round your answers to the nearest hundredth of a second (two decimal places).
Answer: The object is 151 feet off the ground at the following times: __
Enter your two answers separated by a comma.
To find the points in time when the object is 151 feet above the ground, we need to solve the equation h = 151.
-16t^2 + 127t + 114 = 151
Rearranging the equation:
-16t^2 + 127t + 114 - 151 = 0
-16t^2 + 127t - 37 = 0
Using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / 2a
a = -16, b = 127, c = -37
t = (-127 ± sqrt(127^2 - 4(-16)(-37))) / (2(-16))
t = (-127 ± sqrt(16129 - 2368)) / (-32)
t = (-127 ± sqrt(13761)) / (-32)
t = (-127 ± 117) / (-32)
t1 = (-127 + 117) / (-32) = -10 / (-32) = 0.31 (approx)
t2 = (-127 - 117) / (-32) = -244 / (-32) = 7.63 (approx)
Therefore, the object is 151 feet off the ground at the following times: 0.31 seconds and 7.63 seconds.