Is my answer correct?? 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? I got 337.59. Is this correct? Please help.
Your answer is quite close.
I get 337.66.
If you have been copying intermediate answers to 4 figures, that would make the difference. Try to use the calculator memory instead of copying down intermediate answers.
How did you get 337.66, i still get 337.59
Jessica/Sam -- why did you switch names??
Please use the same name for your posts.
Yes, quite close but that's correct. `u`
You can use derivatives and you'll get the 337.66 meters.
To find the maximum height reached by the golf ball, you need to determine the highest point on the trajectory, where the velocity becomes zero. The maximum height occurs when the velocity reaches zero because the ball is momentarily at rest before it starts descending.
To find the maximum height, you'll need to determine the derivative of the function h(t) with respect to time t and solve for when the derivative equals zero.
Let's find the derivative of h(t) first:
h'(t) = 50 - 3.724t
Now, set h'(t) equal to zero and solve for t:
50 - 3.724t = 0
3.724t = 50
t = 50 / 3.724
t ≈ 13.43 seconds
This means that the ball reaches its maximum height at approximately t = 13.43 seconds.
To find the maximum height, substitute this value of t back into the original function h(t):
h(13.43) = 2 + 50(13.43) - 1.862(13.43)^2
Calculating this expression gives you the maximum height reached by the ball. Plugging in the values and simplifying:
h(13.43) ≈ 2 + 673.87 - 1.862(179.5249)
h(13.43) ≈ 2 + 673.87 - 334.56
h(13.43) ≈ 341.31
Therefore, the correct maximum height reached by the golf ball is approximately 341.31 meters, not 337.59 as you had calculated.