Question

An object is thrown upward at a speed of 110 feet per second by a machine from a height of 20 feet off the ground. The height h of the object after t seconds can be found using the equation h = – 16t2 + 110t + 20 When will the height be 59 feet? Select an answer v When will the object reach the ground? Select an answer Question Help: Video Submit Question

Answer 1

The equation that defines the motion is given as:

`h=-16t^2+110t+20`

To find when the height will be 59 feet, substitute h=59 into the equation and solve for t:

`\begin{gathered} 59=-16t^2+110t+20 \\ \text{Swap the sides of the equation:} \\ -16t^2+110t+20=59 \\ -16t^2+110t+20-59=0 \\ -16t^2+110t-39=0 \\ \text{Multiply both sides of the equation by -1:} \\ 16t^2-110t+39=0 \end{gathered}`

Solve the quadratic equation using any method of your choice.

Using the quadratic formula where a=16, b=-110 and c=39:

`\begin{gathered} t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ t=\frac{-(-110)\pm\sqrt[]{(-110)^2-4(16)(39)}}{2(16)} \\ t=\frac{110\pm\sqrt[]{12100-2496}}{32} \\ t=\frac{110\pm\sqrt[]{9604}}{32}_{} \\ t=(110\pm98)/(32) \\ t=(208)/(32)or\text{ }(12)/(32) \\ t=6.5s\text{ or }0.375s \end{gathered}`

The unit of time is seconds.

T

The object will reach the ground when the height is zero, that is when h=0.

Substitute into the equation and solve for t again:

`0=-16t^2+110t+20`

Using the same procedure, where a=-16, b=110 and c=20, use the quadratic formula to solve for t to get:

`t\approx-0.18s\text{ or }t\approx7.05s`

Since time cannot be negative, discard the negative value. So, the object will reach the ground at about t=7.05 seconds.