If an object is propelled upward from a height of 16 ft at an initial velocity of 80 feet per sec, the its height after t seconds is given by the equation h=-16t²+80t+16, where h is in feet. after how many seconds will the object reach a height of 116 ft?
116 = -16 t^2 + 80 t + 16
16 t^2 - 80 t + 100 = 0
4 t^2 - 20 t + 25 = 0
(2 t- 5)^2 = 0
t = 5/2 = 2.5 seconds
To find the time at which the object reaches a height of 116 ft, we need to solve the given equation for h = 116.
The equation representing the height of the object is given by h = -16t² + 80t + 16.
Setting h equal to 116, we have:
116 = -16t² + 80t + 16.
Now, we can rewrite the equation in the standard quadratic form: -16t² + 80t + 16 - 116 = 0.
Combining like terms, we get: -16t² + 80t - 100 = 0.
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b² - 4ac)) / 2a.
Here, a = -16, b = 80, and c = -100.
Substituting the values into the formula, we get:
t = (-80 ± √(80² - 4(-16)(-100))) / (2(-16)).
Simplifying further:
t = (-80 ± √(6400 - 6400)) / (-32).
Since the value inside the square root is zero, it simplifies to:
t = -80 / -32 = 2.5.
Hence, the object will reach a height of 116 ft after 2.5 seconds.