The function h(t)=-16t^2+100t+300 gives the height in feet of a projectile fired from the top of a building in t seconds.
A) When will the object reach a height of 350 feet? Round answer to one decimal place.
B) Explain why part A has two answers.
350 = -16t^2 + 100t + 300
16t^2 - 100t + 50 = 0
t = (100 ±√6800)/32
= 5.70194 or .548
it will reach a height of 350 ft on its way up, and then again on its way down.
To find when the object reaches a height of 350 feet, we need to solve the equation h(t) = 350.
A) To do so, we substitute 350 for h(t) in the equation and solve for t:
-16t^2 + 100t + 300 = 350
Simplifying the equation:
-16t^2 + 100t + 300 - 350 = 0
-16t^2 + 100t - 50 = 0
Now, to solve this quadratic equation, we can either factor it or use the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = -16, b = 100, and c = -50. Substituting these values:
t = (-(100) ± √((100)^2 - 4(-16)(-50))) / (2(-16))
t = (-100 ± √(10000 - 3200)) / (-32)
t = (-100 ± √(6800)) / (-32)
t = (-100 ± 82.462) / (-32)
Now we solve for the two possible values of t:
t1 = (-100 + 82.462) / (-32) ≈ 0.675 seconds (rounded to one decimal place)
t2 = (-100 - 82.462) / (-32) ≈ 5.675 seconds (rounded to one decimal place)
Therefore, the object will reach a height of 350 feet at approximately 0.7 seconds and 5.7 seconds.
B) Part A has two answers because the equation is a quadratic equation, which can have two real solutions in some cases. In this case, the object reaches a height of 350 feet on its ascent to its maximum height and then again on its descent from the maximum height back to 350 feet. The equation h(t) = 350 has two solutions that correspond to these two instances.