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) What is the maximum height reached by the ball (would it be 1.862)
B) How long will the ball be above the surface of Mercury?
To find the maximum height reached by the ball, we need to determine the vertex of the parabolic function h(t) = 2 + 50t - 1.862t^2.
A) To find the maximum height, we can use the formula for the x-coordinate of the vertex of a quadratic function: -b/(2a). In our case, the function is h(t) = -1.862t^2 + 50t + 2, so the coefficient of t^2 is a = -1.862 and the coefficient of t is b = 50.
To find the x-coordinate of the vertex, we can use the formula -b/(2a):
t = -50/(2*(-1.862))
t = 13.46 seconds
To find the maximum height, we substitute this value into the h(t) function:
h(13.46) = 2 + 50(13.46) - 1.862(13.46)^2
h(13.46) ≈ 673.86 meters
Therefore, the maximum height reached by the ball is approximately 673.86 meters.
B) To find how long the ball will be above the surface of Mercury, we need to determine the time intervals for which h(t) is greater than zero (since the ball is above the surface when its height is positive).
We can find the roots of the function h(t) = 2 + 50t - 1.862t^2 by setting it equal to zero and solving for t:
2 + 50t - 1.862t^2 = 0
To solve this quadratic equation, we can use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a). In our case, a = -1.862, b = 50, and c = 2.
Using the quadratic formula, we get:
t = (-50 ± √(50^2 - 4*(-1.862)*2)) / (2*(-1.862))
Simplifying this expression, we find two values for t:
t1 ≈ 0.206 seconds
t2 ≈ 26.93 seconds
The ball will be above the surface of Mercury between these two time intervals: t1 to t2. Therefore, the ball will be above the surface for approximately 26.93-0.206 = 26.724 seconds.