Question

A ball is thrown from an initial height of feet with an initial upward velocity of . The ball's height (in feet) after seconds is given by the following.Find all values of for which the ball's height is feet.Round your answer(s) to the nearest hundredth.(If there is more than one answer, use the "or" button.)

Answer 1

Given the projectile motion as

$\begin{equation*} 2+30t-16t^2 \end{equation*}$

We can find the values of t at h=15 below.

Explanation

At height 15, we will have that;

$15=2+30t-16t^2$

Therefore,

$\begin{gathered} 2+30t-16t^2=15 \\ \mathrm{Subtract\:}15\mathrm{\:from\:both\:sides} \\ 2+30t-16t^2-15=15-15 \\ \mathrm{Simplify} \\ -16t^2+30t-13=0 \\ \end{gathered}$

We can then solve the above with quadratic formula

$\begin{gathered} t_{1,\:2}=\frac{-30\pm\sqrt{30^2-4\left(-16\right)\left(-13\right)}}{2\left(-16\right)} \\ t_{1,\:2}=\frac{-30\pm \:2\sqrt{17}}{2\left(-16\right)} \\ \mathrm{Separate\:the\:solutions} \\ t_1=\frac{-30+2\sqrt{17}}{2\left(-16\right)},\:t_2=\frac{-30-2\sqrt{17}}{2\left(-16\right)} \\ \mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:} \\ t=\frac{15-\sqrt{17}}{16},\:t=\frac{15+\sqrt{17}}{16} \\ t=0.68\:t=1.20 \end{gathered}$

Answer:

$t=0.68;t=1.20$