The function h(t) = 2 + 50t - 1.862t^2, where h(t) is the height in metres and t is the time in seconds, models the height of a golf ball above the planet Mercury's surface during its flight.
A) How long will the ball be above the surface of Mercury? I get t= 22.02 and 4.83. But one of my numbers should be a negative so i know that number cant be time.
I get -0.04 and 26.89
what did you do to get your answers?
I will use derivatives:
h(t) = 2 + 50t - 1.862t^2
dh/dt = 50 - 3.724t
0 = 50 - 3.724t
3.724t = 50
t = 13.426 s
This is the time that the ball reaches its maximum height. But the problem asks for the time the ball is above the surface of Mercury. Therefore we multiply it by 2 to account for the time it falls back the ground:
13.426 * 2 = 26.85 s
Another technique is to let h(t) = 0, because at the ground, that is at position h = 0.
0 = 2 + 50t - 1.862t^2
And solve for t using quadratic equation. You should get the same value (or if not the same, approximately the same).
hope this helps~ `u`
(I apologize if someone has already posted an answer/solution before me. I've got slow internet and I can't seem to post my comment right away.)
To find out how long the golf ball will be above the surface of Mercury, we need to determine when the height function h(t) equals zero. This is because the height of the ball will be zero when it touches the surface of Mercury.
Given the height function h(t) = 2 + 50t - 1.862t^2, we can set it equal to zero and solve for t:
0 = 2 + 50t - 1.862t^2
This equation is a quadratic equation, and 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 our equation 0 = 2 + 50t - 1.862t^2, the coefficients are:
a = -1.862
b = 50
c = 2
Plugging these values into the quadratic formula, we get:
t = (-(50) ± √((50)^2 - 4(-1.862)(2))) / (2(-1.862))
Simplifying further:
t = (-50 ± √(2500 - (-14.496))) / (-3.724)
t = (-50 ± √(2500 + 14.496)) / (-3.724)
t = (-50 ± √(2514.496)) / (-3.724)
Now we can calculate the two possible values for t:
t1 = (-50 + √(2514.496)) / (-3.724)
t2 = (-50 - √(2514.496)) / (-3.724)
Simplifying further:
t1 ≈ 4.831 seconds (rounded to the nearest thousandth)
t2 ≈ -22.015 seconds (rounded to the nearest thousandth)
Now we have two possible values for t: approximately 4.831 seconds and approximately -22.015 seconds.
Since time cannot be negative in this context, we discard the negative value. Therefore, the ball will be above the surface of Mercury for approximately 4.831 seconds.