Question

A

A ball is launched from a 137.2-meter tall
platform. The equation for the ball's height h at
time t seconds after launch is
h(t) = -4.9t2 + 2.94t + 137.2, where h is
in meters. When does the object strike the
ground?
=

Answer 1

The following equation describes the height h of the ball after an amount of time of t seconds.

`h(t)=-4.9t^2+2.94t+137.2`

The ground represents the height zero, therefore, to find the corresponding time when the ball hits the ground we just have to solve the equation

`-4.9t^2+2.94t+137.2=0`

To solve this equation we can use the quadratic formula.

Our solutions are

`\begin{gathered} t_(\pm)=(-(2.94)\pm√((2.94)^2-4(-4.9)(137.2)))/(2(-4.9)) \\ =(-(2.94)\pm√(8.6436+2689.12))/(-9.8) \\ =(2.94\pm√(2697.7636))/(9.8) \\ =(2.94\pm51.94)/(9.8) \end{gathered}`

Our solution is only the positive root.

`t=(2.94+51.94)/(9.8)=5.6`

The object strikes the ground after 5.6 seconds.