A model rocket is shot straight up from the roof of a school. The height at any time t is approximated by the model h=-t^2+3t+28, where h is the height in meters and t is the time in seconds. When does the rocket hit the ground?
it hits the ground when h=0
since h = (-t-4)(t-7) when do you think that is?
To find the time when the rocket hits the ground, we need to determine when the height of the rocket (h) is equal to zero.
Given the equation h = -t^2 + 3t + 28, we can set it equal to zero and solve for t:
0 = -t^2 + 3t + 28
To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula. In this case, let's use the quadratic formula:
The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our equation 0 = -t^2 + 3t + 28, a = -1, b = 3, and c = 28. Substituting these values into the quadratic formula:
t = (-3 ± √(3^2 - 4*(-1)*28)) / (2*(-1))
t = (-3 ± √(9 + 112)) / (-2)
t = (-3 ± √121) / (-2)
Simplifying further:
t = (-3 ± 11) / -2
This gives us two possible solutions:
t1 = (-3 + 11) / -2 = -14 / -2 = 7
t2 = (-3 - 11) / -2 = -14 / -2 = 7
Since time cannot be negative in this context, we discard the negative solution. Therefore, the rocket hits the ground at t = 7 seconds.