Question

When will the height be 71ft? When will the object reach the ground?

Answer 1

Given:

`h=-16t^2+97t+17`

when the height is 71ft,

h = 71

Hence,

`\begin{gathered} 71=-16t^2+97t+17 \\ 16t^2-97t+71-17=0 \\ 16t^2-97t+54=0 \end{gathered}`

Using the quadratic formula;

`\begin{gathered} a=16,b=-97,c=54 \\ t=(-b\pm√(b^2-4ac))/(2a) \\ t=(-(-97)\pm√((-97)^2-(4*16*54)))/(2*16) \end{gathered}`

Hence,

`\begin{gathered} t=(97\pm77.156)/(32) \\ t_1=(97+77.156)/(32)=(174.156)/(32)=5.442375=5.44s \\ t_2=(97-77.156)/(32)=(19.844)/(32)=0.620125=0.62s \end{gathered}`

Therefore, the height will be 71 feet at 5.44 seconds or 0.62 seconds.

The object will reach the ground at;

The object will reach the height of 17ft at;

h = 17

`\begin{gathered} 17=-16t^2+97t+17 \\ 16t^2-97t=0 \\ t(16t-97)=0 \\ t=0 \\ 16t-97=0 \\ 16t=97 \\ t=(97)/(16) \\ t=6.0625s \end{gathered}`

Hence, the time it will take to reach the ground will be the time it takes to go up and return back to the ground;

`\begin{gathered} t=6.0625*2 \\ t=12.125s \end{gathered}`

It will take 12.125 seconds for the object to reach the ground.