A baseball player hit a ball with an upward velocity of 46 ft/s. Its height in feet after t seconds is given by the function h= 16t^2+46t+6. What is the maximu
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2 Answer

1. User Answers sharmaenderp
Given the function h(t) where t is the time, to find the maximum height reached by the ball, we must find the maximum value of h(t) = 16t² + 46t + 6. Since h(t) is a quadratic function, to find its maximum value, we need to look for the coordinates of its vertex (x, y). To solve for the xcoordinate, we have
[tex] x = \frac{b}{2a} [/tex]
[tex] x = \frac{46}{2(16)} = \frac{23}{16} [/tex]
Thus, it takes t = 23/16 s for the ball to reach the maximum height. To find the maximum height, we need to substitute the value of t into h(t).
h = 16(23/16)² + 46(23/16) + 6
h = 625/16 ft
For the ball to reach ground level, then h must be equal to zero. Thus, we have
0 = 16t² + 46t + 6
0 = 4t²  23t  3
Recalling the quadratic formula, we have
[tex] t = \frac{b \pm \sqrt{b^{2}4ac} }{2a} [/tex]
where a , b, and c are the coefficients of the quadratic equation. For our case, a = 4, b = 23, and c = 3. Thus, we have
[tex] t = \frac{23 \pm \sqrt{23^{2}4(4)(23)} }{2(4)} [/tex]
[tex] t = \frac{23 \pm \sqrt{897}}{8} [/tex]
Since time cannot be negative, we disregard the negative root and have t = (23 + √897)/8 s.
Answer:
Maximum time = 23/16 seconds
Maximum height = 625/16 ft
Time taken to reach the ground = (23 + √897)/8 seconds

2. User Answers valeriagonzalez0213
Answer:
maximum height: h=39.06 ft
time for the ball to reach its maximum height: t= 1.4375 s
time it takes for the ball to touch the ground: [tex]t_{g} = 2 * 1.4375 = 2.875s[/tex]
Stepbystep explanation:
The function h (t) has a maximum value when h´(t) = 0
h´(t) is the derivative of h with respect to t.
We find h´(t)) to calculate the time for the ball to reach its maximum height
h = 16t²+ 46t + 6 Equation (1)
h´(t) = 32t + 46
32t + 46 = 0
46 = 32t
t = 46/32 = 1.4375 s
We replace 1.4375 s in equation (1) to calculate the maximum height:
h = 16 (1.4375) ² + 46 (1.4375) + 6
h = 39.06 ft
The time it takes for the ball to touch the ground ([tex]t_{g}[/tex]) is twice the time it takes to reach the maximum height.
[tex]t_{g} =2t[/tex]
[tex]t_{g} = 2 * 1.4375 = 2.875s[/tex]