Some fireworks are fired vertically into the air from the ground at an initial speed of 80 feet per second. The equation for this object’s height h at time t seconds after launch is

h= -16t^2 + 80t
How long will it take the fireworks to reach the ground?

Following are the solution to the given question:

            We'll factor it out.

Therefore, the ball remains just on the ground for a moment, t = 0 (just before you tossed it), so it rises, falls, and hits the ground again 5 seconds later, , that'swhy, the final answer is "5 seconds".

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The firework will take 5 s to reach the ground again.

Step-by-step explanation:

For the firework to reach the ground it's height must be equal to 0, so we need to solve the given equation for t with h=0. We have:

h = -16t² + 80t

0 = -16t² + 80t

16t² -80t = 0

t(16t - 80) = 0

In this case we have an incomplete second order equation, there are only two values that can satisfy that equality, these are given bellow:

t = 0 s

16t - 80 = 0

16t = 80

t = 80/16 = 5 s

Since 0 s is when it was launched, then 5 s is the time it takes to reach the ground again.

f(x) = y

f(-3) = -5

step-by-step explanation:

the first one

step-by-step explanation:

just id the test.