A ball is thrown into the air with an initial upward velocity of 45 ft/s. Its height (h) in feet after t seconds is given by the function h = –16t² + 45t + 6. After about how many seconds will the ball hit the ground? (1 point)
3
4 My ans??
5
6
I get 2.9
how did you get your answer?
Yes, after refiguring it. Thanks
what is the answer ??????
The answer is 3. Use your graphing calculator, enter the equation in the y=, press graph, when the line hits the x-intercept, that's your answer.
It's practically at 3, so the answer is 3. ^^
To determine when the ball hits the ground, we need to find the value of t when the height h is equal to 0.
Given the function h = -16t² + 45t + 6, we set h equal to 0:
0 = -16t² + 45t + 6
Now, we can solve this quadratic equation for t. There are a few methods to do this, but one common approach is to factor or use the quadratic formula.
In this case, let's use the quadratic formula:
t = (-b ± √(b² - 4ac)) / 2a
Here, a = -16, b = 45, and c = 6. Plugging these values into the quadratic formula, we get:
t = (-45 ± √(45² - 4(-16)(6))) / (2(-16))
Now, simplify the equation inside the square root:
t = (-45 ± √(2025 + 384)) / -32
t = (-45 ± √2409) / -32
Using a calculator, you can find that the square root of 2409 is approximately 49.08. So substituting this value into the equation:
t = (-45 ± 49.08) / -32
Now, solve for t separately for the two cases (with both + and -):
Case 1: t = (-45 + 49.08) / -32
Case 2: t = (-45 - 49.08) / -32
Simplifying each case, we get:
Case 1: t ≈ 0.71 seconds
Case 2: t ≈ 4.41 seconds
Since the negative value of time (Case 2) doesn't make sense in this context, we can conclude that the ball will hit the ground after approximately 0.71 seconds.
Therefore, the correct answer is 1) 3 seconds.