Question

If a firework is launched vertically upwards from a height of 20 feet above the ground. Its path is give by h(t)=-4t+16t+20, where h is the height of feet and t is the time in seconds.

can someone help me find the right answers for this with scratch work? i’m struggling quite a bit.
when is the firework at max height?
What was the highest point that the firework reached?
What is the starting point of the firework?
How high is the firework after 3 seconds?
When does the firework hit the ground?

1. thuhuong
• The time of the maximum height is of 2 seconds.
• The maximum height of the firework is of 36 feet.
• The starting point of the firework is of 20 feet.
• After 3 seconds, the firework is at a height of 32 feet.
• The firework hits the ground after 5 seconds.

### What is the quadratic function?

The quadratic function defining the height of the firework in this problem after t seconds is given as follows:
h(t) = -4t² + 16t + 20.
From this definition, we can get the h-intercept giving the starting point of the firework, as follows:
h(0) = -4(0)² + 16(0) + 20 = 20 feet.
The height after 3 seconds is calculated with the numeric value as follows:
h(3) = -4(3)² + 16(3) + 20 = 32 feet.
The coefficients of the quadratic function are given as follows:
a = -4, b = 16, c = 20.
Due to the negative coefficient a, this is a concave down function, meaning that the vertex is a maximum point.
The t-coordinate of the vertex is given as follows:
tv = -b/2a = -16/2(-4) = -16/-8 = 2.
Then the maximum height of the firework is of:
h(2) = -4(2)² + 16(2) + 20 = 36 feet.
The firework hits the ground when h(t) = 0, hence, using a quadratic function calculator, the time is of:
5 seconds. (positive root).