Question

aA ball is launched from a 137.2-meter tallplatform. The equation for the ball's height h attime t seconds after launch ish(t) = -4.9t2 + 2.94t + 137.2, where h isin meters. When does the object strike theground?=

Answer 1

We need to find the value of t for which h(t) = 0.

Notice that a quadratic equation may have two real solutions. Since t starts at 0 and only grows, we are interested in the positive solution only.

We have:

$\begin{gathered} h(x)=0 \\ \\ -4.9t^2+2.94t+137.2=0 \end{gathered}$

Then, using the quadratic equation, we obtain:

$\begin{gathered} t=\frac{-2.94\pm\sqrt{(2.94)^2-4(-4.9)(137.2)}}{2(-4.9)} \\ \\ t=\frac{-2.94\pm\sqrt{8.6436+2689.12}}{-9.8} \\ \\ t=\frac{2.94\pm\sqrt{2697.7636}}{9.8} \\ \\ t=\frac{2.94\pm51.94}{9.8} \\ \\ t_1=\frac{2.94-51.94}{9.8}=-5 \\ \\ t_2=\frac{2.94+51.94}{9.8}=5.6 \end{gathered}$

Answer

The object strikes the ground at: t = 5.6 seconds.