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Nataly_w [17]
3 years ago
9

A firework is launched into the air from ground level with an initial velocity of 128 ft/s. If acceleration due to gravity is –1

6 ft/s2, what is the maximum height reached by the firework?
Mathematics
1 answer:
Hitman42 [59]3 years ago
6 0

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.

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