A model rocket Is launched from a height of 50ft. The formula h=-16t^2+70t +50 describes the rocket's height,h,in feet seconds after it was launched. How long will it take the rocket to reach to reachthe ground?
When it's at the ground h = 0
-16t^2+70t+50 = 0
using the quadratic formula
t=5
t=-.625
t cannot be negative, so t=5s is the correct answer.
http://www.regentsprep.org/regents/math/algtrig/ate3/quadformula.htm
To find out how long it will take for the rocket to reach the ground, we need to determine the value of t when the height, h, is equal to 0.
The given formula is h = -16t^2 + 70t + 50, where h represents the height and t represents time in seconds.
Setting h = 0, we can solve for t using the quadratic equation:
-16t^2 + 70t + 50 = 0
To solve this quadratic equation, we can either factor it or use the quadratic formula. In this case, factoring is not straightforward, so we'll use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In our equation, a = -16, b = 70, and c = 50. Plugging in these values, we have:
t = (-70 ± √(70^2 - 4(-16)(50))) / 2(-16)
Simplifying further:
t = (-70 ± √(4900 + 3200)) / -32
t = (-70 ± √(8100)) / -32
t = (-70 ± 90) / -32
Now we have two possible solutions:
1) t = (-70 + 90) / -32 = 20 / -32 = -0.625
2) t = (-70 - 90) / -32 = -160 / -32 = 5
The time, t, cannot be negative, so we discard the first solution. Therefore, the rocket will reach the ground after 5 seconds.
So, it will take 5 seconds for the rocket to reach the ground.