The height s of a ball, in feet, thrown with an initial velocity of 112 feet per second from an initial height of 20 feet is given as a function of time t (in seconds) by
s(t)= -16t^2+112t+20. What is the maximum height of the ball? At what time does the maximum height occur?
I know how to find the maximum height. You find the vertex which is 3.5 then plug it back into the original equation.
Now, when it says what time, I am not sure if it wants the vertex or should I use the quadratic to solve it?
Please help.
The maximum height is the vertex.
This would be an application of completing the squares to convert the function into the standard form:
s(t)=a(t-h)²+k
where t=h at the vertex (3.5) and
k=maximum height (216 above datum)
Ok thanks.
You're welcome!
To find the maximum height of the ball, you indeed need to find the vertex of the quadratic function.
The formula for the x-coordinate (time) of the vertex of a quadratic function in the form ax^2+bx+c is given by -b/2a.
In this case, the function is s(t) = -16t^2 + 112t + 20, where a = -16 and b = 112.
Using the formula, we can find the time at the maximum height:
t = -b/2a = -112/(2*(-16)) = -112/(-32) = 3.5
So the maximum height occurs at time t = 3.5 seconds.
Next, to find the maximum height of the ball, you can substitute the value of t = 3.5 back into the original equation s(t) = -16t^2 + 112t + 20:
s(3.5) = -16(3.5)^2 + 112(3.5) + 20
s(3.5) = -16(12.25) + 392 + 20
s(3.5) = -196 + 392 + 20
s(3.5) = 216
Therefore, the maximum height of the ball is 216 feet.
In conclusion, the maximum height of the ball is 216 feet, and it occurs at a time of 3.5 seconds.