Question

A rocket is launched from a tower. The height of the rocket, y in teetrelated to the time after launch, x in seconds by the given equation. Usingthis equation, find out the time at which the rocket will reach its mar, to thenearest 100th of a second.y=-16x^2 +218x+ 75

Answer 1

Since it is a quadratic equation: ax^2 + bx + c, we can use a formula for finding the time at which the rocket will reach its max.

`\begin{gathered} Vx=(-b)/(2a) \\ Vx=(-218)/(2(-16)) \\ Vx=(-218)/(-32)(\text{Multiplying)} \\ Vx=6.8125\text{ } \\ \text{The answer would be 6.81 seconds (Rounding to the nearest hundredth of a second)} \end{gathered}`