Question

A ball is thrown from initial height of 2 feet with an initial upward velocity of 35 ft./s. The ball's height H (in feet) after T seconds is given by the following. H=2+35T-16T^2
Find all values of T for which the ball's height is 20 feet. Round your answer(s) to the nearest hundredth.

Answer 1

So if the height is going to be 20ft, that means 20 = 2+35t-16t²

$\bf 20=2+35t-16t^2\implies 16t^2-35t+18=0 \\\\\\ \textit{now, it doesn't factor in integers, thus} \\\\\\ \textit{quadratic formula}\\\\ x= \cfrac{ - {{ b}} \pm \sqrt { {{ b}}^2 -4{{ a}}{{ c}}}}{2{{ a}}}\implies t=\cfrac{-(-35)\pm\sqrt{(-35)^2-4(16)(18)}}{2(16)} \\\\\\ t=\cfrac{35\pm\sqrt{73}}{32}\implies t\approx \begin{cases} 1.361\\ 0.828 \end{cases}$

Answer 2

H=-16t²+35t+2
Solve for h=20
20=-16t²+35t+2
-16t²+35t-18=0
Using the quadratic equation, we get values of 1.36075011704
 and 0.826749882959 as values for t ☺☺☺☺