a model rocket is shot straight up from a roof the height at any time t is aproximated by the model h=-t(squared) +3t+28,where h is the height in metres and t is the time in seconds . When does the rocket hit the ground?
To determine when the rocket hits the ground, you need to find the time at which the height (h) becomes zero.
The given equation for the height of the rocket is h = -t^2 + 3t + 28.
To find when the rocket hits the ground, we set h = 0 and solve for t:
0 = -t^2 + 3t + 28
Now, we have a quadratic equation. To solve it, we can use either factoring, completing the square, or the quadratic formula.
In this case, let's solve it by factoring:
0 = -t^2 + 3t + 28
Rearranging the terms:
t^2 - 3t - 28 = 0
Now, we factor the equation:
(t - 7)(t + 4) = 0
This gives us two possible solutions:
t - 7 = 0 --> t = 7 seconds
t + 4 = 0 --> t = -4 seconds
Since time cannot be negative in this context, the rocket hits the ground at t = 7 seconds.
Therefore, the rocket hits the ground after 7 seconds.
To find when the rocket hits the ground, we need to determine the value of "t" when the height "h" is equal to zero.
Given the equation: h = -t^2 + 3t + 28
Setting h = 0, we can solve for t:
0 = -t^2 + 3t + 28
Rearranging the quadratic equation to standard form:
t^2 - 3t - 28 = 0
To solve this quadratic equation, we can factorize or use the quadratic formula.
Factoring:
(t - 7)(t + 4) = 0
Setting each factor to zero:
t - 7 = 0 or t + 4 = 0
Solving for t in each case:
t = 7 or t = -4
Since time cannot be negative in this context, the rocket hits the ground when t = 7 seconds.
hit the ground? then h=0
0=-t^2 + 3t+28=t^2-3t-28=(t-7)(t+4)=0
t=7 seconds