# Max Height of Ball

• July 21st 2008, 04:58 AM
magentarita
Max Height of Ball
The height s of a ball (in feet) thrown with an initial velocity of 60 feet per second from an initial height of 6 feet is given as a function of the time t (in seconds) by s(t) = (-16t^2) + 60t + 6

(a) Determine the time at which height is maximum.

(b) What is the maximum height?

NOTE: I have trouble with questions like this all the time. I never know what number(s) to plug into the formula.
• July 21st 2008, 05:16 AM
Simplicity
Quote:

Originally Posted by magentarita
The height s of a ball (in feet) thrown with an initial velocity of 60 feet per second from an initial height of 6 feet is given as a function of the time t (in seconds) by s(t) = (-16t^2) + 60t + 6

(a) Determine the time at which height is maximum.

(b) What is the maximum height?

NOTE: I have trouble with questions like this all the time. I never know what number(s) to plug into the formula.

(a)
$s = -16t^2 + 60t + 6$.

Find $\frac{\mathrm{d}s}{\mathrm{d}t}$ and equate this to zero. You then rearrange to find the time by making $t$ the subject.

(b)
Substitute the value of time (found in question (a)) into the function ( $s = -16t^2 + 60t + 6$) to find the maximum height.

EDIT: Maximum & Minimum of Graph <This weblink explain the theory and provided a few question for you to try out.
• July 21st 2008, 06:00 AM
Soroban
Hello, magentarita!

I assume this is Pre-Calculus . . .

Quote:

The height $s$ of a ball (in feet) thrown with an initial velocity of 60 ft/sec
from an initial height of 6 ft is given by: . $s(t) \:= \:-16t^2 + 60t + 6$

(a) Determine the time at which height is maximum.

The height function is: . $s \:=\:-16t^2 + 60t + 6$

This is a down-opening parabola.
. . Its maximum is at its vertex.

The vertex is: . $t \:=\:\frac{-b}{2a} \:=\:\frac{-60}{2(-16)} \:=\:\frac{15}{8}$

The ball reaches maximum height in $\boxed{1.875\text{ seconds}}$

Quote:

(b) What is the maximum height?
$s\left(\frac{15}{8}\right) \:=\:-16\left(\frac{15}{8}\right)^2 + 60\left(\frac{15}{8}\right) + 6 \;=\;\frac{249}{4}$

. . The maximum height is: . $\boxed{62.25\text{ feet}}$

• July 21st 2008, 06:32 AM
magentarita
Soroban...
Yes, this is a precalculus question. This is the reason why I placed it in this category. Your explanation is easy to follow and completely cleared any confusion in terms of such questions. Are questions like this one often found in physics textbooks? My friend said yes.