Question 1072632
.
<pre>
The maximal height is the vertex of the parabola

h(t) = {{{-16t^2 + 96t}}}

Its t-value is t = {{{-96/(2*(-16))}}} = {{{96/32}}} = 3.

So, the maximal height is achieved at t = 3 seconds.

The value of the maximal height is {{{h[max]}}} = {{{-16*3^2 + 96*3}}} = 144 ft.

<U>Answer</U>. The maximal height is 144 ft at t = 3 seconds.
</pre>


{{{graph( 330, 330, -1.5, 8.5, -20.5, 200.5,
          -16x^2 + 96x
)}}}


Plot y = -16*t^2 + 96t



See the lessons on a projectile thrown/shot/launched vertically up

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/travel/Problem-on-projectile-shooted-vertically-upward.lesson>Problem on an arrow shot vertically upward</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/travel/Typical-problems-on-an-projectile-moving-vertically-up-and-down.lesson>Problem on a ball thrown vertically up from the top of a tower</A> 


Also, you have this free of charge online textbook in ALGEBRA-I in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>.


The referred lessons are the part of this textbook under the topic "<U>Projectiles launched/thrown and moving vertically up and dawn</U>".