Question

An object is launched at 18.4 meters per second (m/s) from a 36.8-meter tall platform. The equation for the object's heights at time t seconds after launch is s(t) = -4.912 + 18.4t + 36.8, where s is in meters. • When does the object strike the ground? (Select ] How long did it take the object to get to its maximum height? Select ] What was the height of the object at 3.32 seconds? | Select ]

Answer 1

`s(t)=-4.912t^2+18.4t+36.8`

1) To find when the object strikes the ground, we need to find the roots of the equation. Using quadratic formula:

`\begin{gathered} t_(1,2)=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ t_(1,2)=\frac{-18.4\pm\sqrt[]{18.4^2-4(-4.912)(36.8)}}{2(-4.912)} \\ t_(1,2)=\frac{-18.4\pm\sqrt[]{1061.6064}}{-9.824} \\ \\ t_1=(-18.4+32.582)/(-9.824)=-1.44 \\ t_2=(-18.4-32.582)/(-9.824)=5.2 \end{gathered}`

t can't be negative, then the object strikes the ground after 5.2 seconds

2) The maximum height is the vertex of the parabola. The t-coordinate is computed as follows:

`t=(-b)/(2a)=(-18.4)/(2(-4.912))=1.87`

It takes 1.87 seconds for the object to get to its maximum height

3) To find the height after 3.32 seconds, we have to replace t = 3.32 int the equation:

`\begin{gathered} s(3.32)=-4.192(3.32)^2+18.4(3.32)+36.8 \\ s(3.32)=-54.142+61.088+36.8 \\ s(3.32)=43.746 \end{gathered}`

The height was 43.746 meters