For a model rocket, the altitude h in meters, as a function of time t, in seconds, is given by h=68t-8t^2. after how many seconds will it hit the ground?

When it is at ground level, h(t)=0.

So solve for h(t)=0,
h(0)=68t-8t^2
=t(68-8t)=0
t=0 or t=68/8=17/4=4.25

To find the time when the model rocket hits the ground, we need to determine the value of t when the altitude h is equal to zero.

Given that the altitude is given by the equation h = 68t - 8t^2, we set h = 0 and solve for t:

0 = 68t - 8t^2

Now we have a quadratic equation, so we can solve it by factoring, completing the square, or using the quadratic formula. In this case, let's use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

For this equation, a = -8, b = 68, and c = 0. Plugging these values into the quadratic formula, we have:

t = (-68 ± √(68^2 - 4(-8)(0))) / (2(-8))
t = (-68 ± √(4624)) / (-16)
t = (-68 ± 68) / (-16)

Now we have two possible solutions:

t1 = (-68 + 68) / (-16) = 0 / (-16) = 0
t2 = (-68 - 68) / (-16) = -136 / (-16) = 8.5

Since time cannot be negative in this context, we discard the negative value of t and conclude that the model rocket will hit the ground after 8.5 seconds.

To find the time at which the rocket hits the ground, we need to determine when the altitude (h) is equal to zero.

The given equation for the altitude of the rocket is: h = 68t - 8t^2

Setting h to zero, we get: 0 = 68t - 8t^2

To solve this quadratic equation, we can rearrange it as: 8t^2 - 68t = 0

Now, we factor out the common factor of 4t: 4t(2t - 17) = 0

From this equation, we can see that either 4t = 0 or 2t - 17 = 0

Solving for each case, we have:
1) 4t = 0
t = 0

2) 2t - 17 = 0
2t = 17
t = 8.5

So, the rocket will hit the ground at two times: t = 0 seconds (initial launch) and t = 8.5 seconds.