The height s of a ball, in feet, thrown with an initial velocity of 112 feet per second from an initial height of 6 feet is given as a function of time t (in seconds) by
s(t)=-16t^2+112t+20.
1. What is the maximum height of the ball?
2. At what time does the maximum height occur?
I know how to find the maximum height using the formula -b/2a and I got 3.5. Usually when I find the maximum height I would plug 3.5 into the original equation. The problem says initial height of 6 feet which is not in the equation presented to me so I'm confused about how to solve for the time if the 6 feet is supposed to be somewhere.
oops! don't know why I put AP Calc, it's actually pre-calc....
I am baffled by the same question. You may want to double check the numbers, or your instructor could help you.
As you suggested, if there is no typo, then either the reference point for the height is 14 feet below the mentioned datum, or the time reference is shifted.
Thanks for the info. I will ask my teacher on Monday.
You're welcome.
I hope your teacher will clear this up on Monday.
To solve for the maximum height of the ball, you are correct in using the formula -b/2a. However, in this case, we need to take into consideration the initial height of the ball as well.
Let's start by rewriting the equation for the height of the ball as a function of time:
s(t) = -16t^2 + 112t + 6
The "6" represents the initial height of the ball. Now, to find the maximum height, we need to find the vertex of the parabolic function. The x-coordinate of the vertex represents the time at which the maximum height occurs, and the y-coordinate represents the maximum height itself.
First, let's rewrite the equation in standard form to easily identify the coefficients:
s(t) = -16t^2 + 112t + 6
Now, we can see that a = -16, b = 112, and c = 6. Substituting these values into the formula -b/2a, we can find the time at which the maximum height occurs:
t = -112 / (2 * -16)
t = -112 / -32
t = 3.5
Therefore, the maximum height occurs at t = 3.5 seconds.
To find the maximum height, we now need to substitute this value of t back into the original equation for s(t):
s(3.5) = -16(3.5)^2 + 112(3.5) + 6
s(3.5) = -16(12.25) + 392 + 6
s(3.5) = -196 + 392 + 6
s(3.5) = 202
Therefore, the maximum height of the ball is 202 feet.