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melomori [17]
3 years ago
14

A water rocket is launched froni a platform. The height,

Mathematics
1 answer:
Alex73 [517]3 years ago
7 0

Answer:

4.41seconds later

Step-by-step explanation:

Let the equation of the height reached by the water front be h = -2t^2+7t+4

The water hits the ground at h =0

Substitute

0 =  -2t^2+7t+4

-2t^2+7t+4 = 0

2t^2 - 7t - 8 = 0

t = 7±√(-7)²-4(2)(-8)/2(2)

t = 7±√49+64/4

t = 7±√113/4

t = 7±10.63/4

t = 7+10.63/4

t = 17.3/4

t = 4.41s

Hence the water hit the ground 4.41seconda later

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