A toy rocket is launched into the air at an initial velocity of 32 ft/sec, as shown on the graph below. The function s(t)=-16t^2+32t+128 gives the height of the rocket (in feet) at time t (seconds). When does the rocket hit the ground? Show your work.
the ground is zero feet
0 = -16t^2 + 32t + 128
dividing by -16 ... 0 = t^2 - 2t - 8
factoring ... 0 = (t - 4) (t + 2)
0 = t - 4 ... t = 4
To find when the rocket hits the ground, we need to determine the time at which the height is equal to zero.
Given the function s(t) = -16t^2 + 32t + 128, we can set it equal to zero and solve for t:
-16t^2 + 32t + 128 = 0
We can simplify this equation by dividing through by -16:
t^2 - 2t - 8 = 0
Now we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. In this case, using factoring:
(t - 4)(t + 2) = 0
This equation gives us two possible solutions: t - 4 = 0 or t + 2 = 0.
Solving for each:
t - 4 = 0
t = 4
t + 2 = 0
t = -2
However, since time cannot be negative, we discard the solution t = -2.
Therefore, the rocket hits the ground after 4 seconds.
To find when the rocket hits the ground, we need to determine the value of t when the height, given by the function s(t), is equal to 0.
The function s(t) represents the height of the rocket at time t and is given by s(t) = -16t^2 + 32t + 128.
Setting s(t) equal to 0:
-16t^2 + 32t + 128 = 0
Next, we can solve this quadratic equation using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = -16, b = 32, and c = 128.
Plugging in these values:
t = (-(32) ± √((32)^2 - 4(-16)(128))) / (2(-16))
Simplifying further:
t = (-32 ± √(1024 + 8192)) / (-32)
t = (-32 ± √(9216)) / (-32)
t = (-32 ± 96) / (-32)
Now, we have two possible solutions:
t1 = (-32 + 96) / (-32)
t1 = 64 / -32
t1 = -2
t2 = (-32 - 96) / (-32)
t2 = -128 / -32
t2 = 4
Since time cannot be negative in this context, we conclude that the rocket hits the ground at t = 4 seconds.
Therefore, the rocket hits the ground after 4 seconds of being launched.