Question

The company will be launching a firework from the ground, but you want to increase the time it stays in the air by 2 seconds to build up the suspense. How can the company make this happen without changing the length of the fuse? How will this affect the maximum height of the firework? g(t) = -16t2 + 224t + 120

Answer 1

Answer: find the answer in the explanation.

Step-by-step explanation:

To increase the time the fireworks stays in the air by 2 seconds to build up the suspense, 2 should be added to the time t.

T + 2 = t

T = t - 2

Substitute T for t in the function g(t)

The company will make this happen without changing the length of the fuse by substituting T for t which will lead to

g(t) = -16(t - 2)^2 + 224(t - 2) + 120

How will this affect the maximum height of the firework?

Initially the function is:

g(t) = -16t2 + 224t + 120

Where the time at maximum height is found by using the formula

t = -b/2a

b = 224, a = -16

Substitutes both into the formula

t = -224/2(-16)

t = -224/-32

t = 7

Substitute 7 for t in the first equation

g(t) = -16(7)^2 + 224(7) + 120

g(t) = -16(49) + 224(7) + 120

g(t) = -784 + 1568 + 120

g(t) = 904 metres

But when the time is delayed by 2 seconds in the air, the maximum height will be

g(t) = -16( 7 - 2 )^2 + 224(7 - 2) + 120

g(t) = -16(5)^5 + 224(5) + 120

g(t) = -16(25) + 224(5) + 120

g(t) = -400 + 1120 + 120

g(t) = 840

The effect will surely reduce the maximum height of the fireworks.