An Object Is Launched At 19.6 Meters Per Second (M/S) From A 58.8-Meter Tall Platform. The Equation

An object is launched at 19.6 meters per second (m/s) from a 58.8-meter tall platform. The equation for the objects height s at time t seconds after launch is s(t) = -4.9t² + 19.6t + 58.8, where s is in meters.

a) What is the maximum height of the object?

 

Step-by-step explanation:

\begin{gathered}\begin{gathered}\sf{By \: the \: time \: of \: the \: object \: launch \: the \: ground,s(t) = 0.} \\ { \sf Thus, solving \: for \: t} \\ \boxed{\begin{array}{l} s(t) = -4.9t^{2} + 19.6t + 58.8 \\ \: \: \: 0 = -4.9t^{2} + 19.6t + 58.8 \\ \: \: \: 0 = -4.9(t^{2} - 4t - 12) \\ \: \: \: 0 = -4.9(t - 6)(t + 2) \\ \: \: \: \: \: \: \: \: \: t = 6 \: or \: t = -2 \end{array}}\end{gathered}\end{gathered}

So, The height maximum of the object is 6 seconds.

Answer : \boxed{\rm 6 \: seconds}


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