Solved on Feb 16, 2024

Solve for t in the equation h=7+35t16t2h = 7 + 35t - 16t^2, given that h=25h = 25.

STEP 1

Assumptions
1. The equation given is h=7+35t16t2 h = 7 + 35t - 16t^2 .
2. We are given h=25 h = 25 .
3. We need to solve for t t .

STEP 2

Substitute h=25 h = 25 into the equation.
25=7+35t16t225 = 7 + 35t - 16t^2

STEP 3

Rearrange the equation to set it to zero, which is the standard form for a quadratic equation.
16t235t+(725)=016t^2 - 35t + (7 - 25) = 0

STEP 4

Simplify the constant terms.
16t235t18=016t^2 - 35t - 18 = 0

STEP 5

We now have a quadratic equation in the form at2+bt+c=0 at^2 + bt + c = 0 , where a=16 a = 16 , b=35 b = -35 , and c=18 c = -18 . We can solve this equation using the quadratic formula:
t=b±b24ac2at = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

STEP 6

Substitute a=16 a = 16 , b=35 b = -35 , and c=18 c = -18 into the quadratic formula.
t=(35)±(35)24(16)(18)2(16)t = \frac{-(-35) \pm \sqrt{(-35)^2 - 4(16)(-18)}}{2(16)}

STEP 7

Simplify the equation by calculating inside the square root.
t=35±1225+115232t = \frac{35 \pm \sqrt{1225 + 1152}}{32}

STEP 8

Add the values inside the square root.
t=35±237732t = \frac{35 \pm \sqrt{2377}}{32}

STEP 9

Calculate the square root.
t=35±237732t = \frac{35 \pm \sqrt{2377}}{32}

STEP 10

Since the square root of 2377 is not a perfect square, we will keep it under the square root sign. We now have two possible solutions for t t :
t=35+237732ort=35237732t = \frac{35 + \sqrt{2377}}{32} \quad \text{or} \quad t = \frac{35 - \sqrt{2377}}{32}

STEP 11

Calculate the approximate values of t t for both solutions.
t35+48.7632ort3548.7632t \approx \frac{35 + 48.76}{32} \quad \text{or} \quad t \approx \frac{35 - 48.76}{32}

STEP 12

Calculate the approximate values for both solutions.
t83.7632ort13.7632t \approx \frac{83.76}{32} \quad \text{or} \quad t \approx \frac{-13.76}{32}

STEP 13

Simplify the fractions to find the approximate values of t t .
t2.6175ort0.43t \approx 2.6175 \quad \text{or} \quad t \approx -0.43
Since time cannot be negative in this context, we discard the negative solution.
The solution for t t when h=25 h = 25 is approximately t2.6175 t \approx 2.6175 .

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