Hard quadratic equation math problem?? Help please?

Jennifer teed off her golf ball from the first tee. and it followed the trajectory given by the function: h(t) = -16t^2 + 100t + 0.25, where t is the time in seconds, and h is the height of the ball above the ground in feet. Find the highest point that her golf ball reached and also when it hits the ground again. Find a reasonable domain and range for this situation. Explain how you know for each.

a. What was the height above ground of the golf ball just before Jennifer teed off?

b. How long was the golf ball in flight?

c. At 1.5 seconds, the ball was approximately 100' above the ground. At what other time is the ball at the same height?

d. At what time does the ball reach the peak of the flight?

e. How high does the ball reach at that time? 

Your help means a lot, thank you in advance. 

3 Answers

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  • 1 month ago

    You have good answers for a-e.  You were also asked for a reasonable domain and range.

    Domain: [0 , 6.25]. This is the time from when the ball is hit until it reaches the ground.  Time before the ball was hit (t < 0) does not fit the curve.  For some time before the ball was struck, it was stationary on the tee.  Time after the ball reaches the ground (t > 6.25) also does not fit the curve.  The ball does not continue to go down, faster and faster because of gravity.  Instead it probably bounces or rolls.

    Range [0 , 156.25]. The domain determines the reasonable range.

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  • ted s
    Lv 7
    1 month ago

    even if you are only in the 9th grade studying parabolas y = a x² + bx + c YOU should know that c is the initial height , - b / 2a is the  domain value for  max height & y(-b / 2a) yields that height , ...x =  (-b/2a) is a line of symmetry so y(w) ≡ y( (- b / a) - w)...{ w = 1.5 in your query}.....and time to landing  is when y = 0 or x = [ - b ± √( b² - 4ac) ] / [ 2 a ].....do the work

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  • 1 month ago

    You are given:

    h(t) = -16t² + 100t + 0.25

    This means that the ball was hit with an initial velocity of 100 ft/s and the tee was 0.25 ft off the ground. (which sounds like to be a pretty tall tee, that's 3 inches!)

    So 0.25 ft (or 3 in) is the answer to part a.

    For part b, how long was the ball in flight, solve for the positive "t" value when h(t) = 0 (when the height is 0, it's on the ground).

    0 = -16t² + 100t + 0.25

    I'll start by dividing both sides by -16:

    0 = t² - (25/4)t - 4

    I'll use complete the square:

    4 = t² - (25/4)t

    Half of -25/4 is -25/8.

    The square of -25/8 is 625/64.

    Add that to both sides:

    4 + 625/64 = t² - (25/4)t + 625/64

    Simplify the left side, factor the right:

    256/64 + 625/64 = (t - 25/8)²

    881/64 = (t - 25/8)²

    The square root of both sides:

    ± √(881)/8 = t - 25/8

    Add 25/8 to both sides:

    t = 25/8 ± √(881)/8

    t = (25 ± √881) / 8

    The smaller root is negative, so throw that out:

    t = (25 + √881) / 8

    t ≈ 6.835 seconds

    That's the time the ball was in the air after being struck

    • llaffer
      Lv 7
      1 month agoReport

      Thanks. I fixed it.

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