Question

A ball is thrown from a height of 32 meters with an initial downward velocity of 2 m/s. The ball's height h (in meters) after t seconds is given by the following.h = 32 -21-57How long after the ball is thrown does it hit the ground?Round your answer(s) to the nearest hundredth.(If there is more than one answer, use the "or" button.)

Answer 1

The given expression is

`h=32-2t-5t^2`

The ground is at h = 0, so

`0=32-2t-5t^2`

Now, we use the quadratic formula to find the solutions.

`x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

Where a = -5, b = -2, and c = 32.

`\begin{gathered} x=\frac{-(-2)\pm\sqrt[]{(-2)^2-4\cdot(-5)\cdot(32)}}{2\cdot(-5)}=\frac{2\pm\sqrt[]{4+640}}{-10}=\frac{2\pm\sqrt[]{644}}{-10} \\ x=(2\pm25.38)/(-10) \\ x_1=(2+25.38)/(-10)=-2.7 \\ x_2=(2-25.38)/(-10)=2.34 \end{gathered}`

In this case, just the positive solution makes sense.

Therefore, the ball takes 2.34 seconds to hit the ground.