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)
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I get 2.9, so rounding it, it would be 3 which i guess is the answer?
or is it 4? (that was my original answer before I redid it and switched it to 3, confused...)
I got 2.9 as well. Your answer is correct =)
To find out how many seconds it will take for the ball to hit the ground, we need to determine when the height (h) of the ball is equal to zero.
The given equation for the height of the ball is h = -16t^2 + 45t + 6, where h represents the height in feet and t represents the time in seconds.
To solve the equation for h = 0, we substitute zero for h:
0 = -16t^2 + 45t + 6
Now, we can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. In this case, factoring may not be easy or possible, so we will use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = -16, b = 45, and c = 6.
Substituting these values into the formula, we get:
t = (-45 ± √(45^2 - 4 * -16 * 6)) / (2 * -16)
Simplifying this expression further, we have:
t = (-45 ± √(2025 + 384)) / (-32)
t = (-45 ± √(2409)) / (-32)
Now that we have the discriminant (√(2409)), we can approximate its value:
√(2409) ≈ 49.08
So, the equation becomes:
t = (-45 ± 49.08) / (-32)
Now, we have two possible solutions:
t1 = (-45 + 49.08) / (-32) ≈ 0.13
t2 = (-45 - 49.08) / (-32) ≈ 3.03
So, rounding to the nearest whole number, the ball will hit the ground after approximately 3 seconds. Therefore, the answer is 3.