Question

A firework is launched into the air from ground level with an initial velocity of 128 ft/s. If acceleration due to gravity is –16 ft/s2, what is the maximum height reached by the firework?

Answer 1

Answer:

h = 256 feet

Step-by-step explanation:

Given that,

The initial velocity of firework, v = 128 ft/s

The acceleration due to gravity is-16 ft/s².

We need to find the maximum height reached by the firework.

The equation for the firework is :

$h(t) = -16t^2 + 128t$

To find the vertex's x-coordinate, we can use  

$t=\dfrac{-b}{2a}\\\\t=\dfrac{-128}{2\times (-16)}\\\\t=4\ s$

Put all the values in the expression for h (t).

$h(t) = -16(4)^2 + 128(4)\\\\h(t)=256\ ft$

So, the maximum height reached by the firework is 256 feet.