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Liam launches a toy rocket from a platform. The height of the rocket in feet is given by h(t)=-16t^(2)+40 t+96 where t represents the time in seconds after launch. How many seconds have gone by when the rocket is at its highest point? Answer: seconds

Liam launches a toy rocket from a platform. The height of the rocket in feet is given by h(t)=−16t2+40t+96 h(t)=-16 t^{2}+40 t+96 where t t represents the time in seconds after launch. How many seconds have gone by when the rocket is at its highest point?\newlineAnswer: seconds

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Q. Liam launches a toy rocket from a platform. The height of the rocket in feet is given by h(t)=−16t2+40t+96 h(t)=-16 t^{2}+40 t+96 where t t represents the time in seconds after launch. How many seconds have gone by when the rocket is at its highest point?\newlineAnswer: seconds
  1. Identify values of aa, bb, cc: Identify the values of aa, bb, and cc in the quadratic equation h(t)=−16t2+40t+96h(t) = -16t^2 + 40t + 96. Compare h(t)=−16t2+40t+96h(t) = -16t^2 + 40t + 96 with the standard form of a quadratic equation, which is at2+bt+cat^2 + bt + c. a=−16a = -16 bb00 bb11
  2. Find time of maximum height: Use the formula for the xx-coordinate of the vertex of a parabola, which is given by −b2a-\frac{b}{2a}, to find the time at which the rocket reaches its maximum height.\newlineSubstitute a=−16a = -16 and b=40b = 40 into the formula.\newlinet=−402(−16)t = -\frac{40}{2(-16)}\newlinet=−40−32t = -\frac{40}{-32}\newlinet=4032t = \frac{40}{32}\newlinet=1.25t = 1.25
  3. Round to nearest hundredth: Round the result to the nearest hundredth of a second to find the time at which the rocket reaches its maximum height.\newlinet=1.25t = 1.25 seconds (This value is already to the nearest hundredth, so no further rounding is necessary.)

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