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patriot [66]
2 years ago
13

A fireworks display is launched from a platform 10 feet above ground with an initial upward velocity of 70 feet per second. The

height of the fireworks above ground after t seconds is given by the equation h = –16t2 + 70t + 10, where h is the height of the fireworks in feet and t is the time in seconds after they are launched. What is the maximum height of the fireworks display, to the nearest foot?

Mathematics
1 answer:
Lena [83]2 years ago
6 0

Answer: A. 87 feet


Step-by-step explanation:

1. You can find the t value of the vertex of the parabola as following:

t=-\frac{b}{2a}

2. Substitute values:

a=-16

b=70

Then:

 t=-\frac{70}{2(-16}

t=2.18

3. Substitute the vaue obtained into the equation given in the problem. Therefore, you obtain the following result:

h=-16t^{2}+70t+10\\h=-16(2.18)^{2}+70(2.18)+10\\h=86.56

4. To the nearest foot:

h=87 feet

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