A toy rocket is launched form a 3m platform, at 8.2m/s. The height of the rocket is modelled by the equation h=-4.0t^2+8.1t+3 where h is the height, in metres, above the ground and t is the time, in seconds.

After how many seconds will the rocket rise to a height of 6m above the ground?

To find the time it takes for the rocket to rise to a height of 6m above the ground, we need to solve the equation for t when h = 6.

Given equation: h = -4.0t^2 + 8.1t + 3

We substitute h = 6 into the equation:
6 = -4.0t^2 + 8.1t + 3

Now we rearrange this equation to solve for t. Moving all terms to one side, we get:
-4.0t^2 + 8.1t - 3 = 0

To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = -4.0, b = 8.1, and c = -3.

Substituting these values into the quadratic formula:
t = (-8.1 ± √(8.1^2 - 4*(-4.0)*(-3))) / (2*(-4.0))

Calculating the expression inside the square root:
t = (-8.1 ± √(65.61 - 48)) / (-8.0)

Simplifying further:
t = (-8.1 ± √(17.61)) / (-8.0)
t = (-8.1 ± 4.2) / (-8.0)

We have two possible solutions:
t1 = (-8.1 + 4.2) / (-8.0)
t2 = (-8.1 - 4.2) / (-8.0)

Calculating each solution separately:
t1 = (-3.9) / (-8.0)
t1 ≈ 0.48875 seconds

t2 = (-12.3) / (-8.0)
t2 ≈ 1.5375 seconds

Since time cannot be negative in this context, we discard t2 as an extraneous solution.

Therefore, the rocket will rise to a height of 6m above the ground after approximately 0.489 seconds.

this was answered when you were Rachel.