If an object is dropped from a height of 126 feet, the function h(t)=-16t^2+126 gives the height of the object after t seconds. Approximately when will the object hit the ground?
solve for t when h=0.
16t^2 = 126
...
set the function to 0 and solve. so it would be 0 = -16t^2+126 and solve for t.
To find approximately when the object will hit the ground, we need to determine the value of t when the height h(t) is equal to zero. This is because when the object hits the ground, its height will be zero.
Given the function h(t) = -16t^2 + 126, we can set h(t) = 0 and solve for t:
-16t^2 + 126 = 0
To solve this quadratic equation, we can factor it or use the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -16, b = 0, and c = 126. Substituting these values into the formula:
t = (0 ± √(0^2 - 4(-16)(126))) / (2(-16))
= ± √(0 - (-8064)) / (-32)
= ± √(8064) / (-32)
≈ ± 28.4 / (-32)
Approximately, t ≈ ± 0.8875
We discard the negative solution since time cannot be negative in this context. Therefore, the object will hit the ground approximately at t ≈ 0.8875 seconds.
Note: This is an approximation and assumes no factors such as air resistance or other external forces affecting the motion of the object.