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.
What is the maximum height reached by the ball? My answer is 341.35
How long will the ball be above the surface of Mercury? My answer is 13.5 seconds
Are my answers correct?
Max height - yes, we got approximately the same answer
Time - the time you got is the time where the ball reached the maximum height. The question asks for the total time (the time the ball is above the surface of Mercury), therefore you should multiply it by 2, so it would account for the time it falls back on the ground.
Hope this helps~ `u`
To determine if your answers are correct, we need to find the maximum height reached by the ball and the time it stays above the surface of Mercury.
To find the maximum height, we can take the derivative of the function h(t) and set it equal to zero. This will give us the value of t at which the ball reaches its maximum height.
Let's calculate the derivative:
h'(t) = 50 - 2 * 1.862t
Setting h'(t) equal to zero:
50 - 2 * 1.862t = 0
Solving for t:
50 = 3.724t
t = 50 / 3.724 ≈ 13.416 seconds
Now, to find the maximum height, we can substitute this value of t into the function h(t):
h(13.416) = 2 + 50 * 13.416 - 1.862 * (13.416)^2
h(13.416) ≈ 341.36 meters
Therefore, the maximum height reached by the ball is approximately 341.36 meters.
As for the time it stays above the surface of Mercury, we can consider the range of values for t where h(t) is greater than zero. This represents the time the ball spends above the surface.
To determine this range, we set h(t) greater than zero:
2 + 50t - 1.862t^2 > 0
Solving this inequality will give us the range of t values when the ball is above the surface.
We can factor this quadratic inequality as follows:
-1.862t^2 + 50t + 2 > 0
This quadratic has roots, t = (-50 ± √(50^2 - 4 * (-1.862) * 2)) / (2 * -1.862)
Simplifying the expression:
t ≈ 0.0404 seconds or t ≈ 26.787 seconds
So, the ball is above the surface of Mercury for approximately t = 26.787 seconds - 0.0404 seconds ≈ 26.7476 seconds.
Therefore, your answer of 13.5 seconds for the time the ball stays above the surface of Mercury is incorrect.
To summarize:
- The maximum height reached by the ball is approximately 341.36 meters.
- The ball stays above the surface of Mercury for approximately 26.7476 seconds.
Please note that the actual maximum height and time may vary slightly depending on the precise calculation method used and the accuracy of the input values.
To find the maximum height reached by the golf ball, we need to find the vertex of the function. The vertex is the point (t, h) where the parabolic function reaches its highest or lowest point. In this case, the maximum height corresponds to the highest point.
To find the vertex, we can use the formula: t = -b / (2a)
Given that the function is h(t) = 2 + 50t - 1.862t^2, we can see that a = -1.862 and b = 50.
Plugging these values into the formula, we have:
t = -50 / (2 * -1.862)
t = 50 / 3.724
t ≈ 13.44
So, the ball will reach its maximum height at approximately 13.44 seconds, not 13.5 seconds. Therefore, your answer is incorrect.
To find the maximum height reached by the ball, we substitute the value of t = 13.44 into the original equation:
h(13.44) = 2 + 50(13.44) - 1.862(13.44)^2
Plugging this into a calculator, we find:
h(13.44) ≈ 341.39
So, the correct maximum height reached by the ball is approximately 341.39 meters. Therefore, your answer of 341.35 meters is very close, but not entirely accurate.
In summary, the correct answers are:
- The maximum height reached by the ball is approximately 341.39 meters.
- The ball will be above the surface of Mercury for approximately 13.44 seconds.