Question

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?

Answer 1

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

Answer 2

Answer:

87 feet

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

2. Substitute values:

a=-16

b=70

Then:

 

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

4. To the nearest foot:

h=87 feet

Step-by-step explanation: