A water rocket is launched froni a platform. The height, h in metres, of the water rocket at time & seconds after
it is launched is h= -212 + 74 + 4. When does the water
rocket hit the ground?​

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

d

step-by-step explanation:

i used the pitagoras theoreme