A model rocket is launched from the ground with an initial velocity of 40 feet per second. The height of an object h, in feet, after t seconds, with initial velocity v0v0 and initial height h0h0 is given by h(t)=−16t2+v0t+h0h(t)=−16t2+v0t+h0 .What is the approximate maximum height the rocket reaches?Enter your answer in the box. ft
Accepted Solution
A:
The rocket reaches a maximum height of 25 feet.
We substitute the velocity given into the equation: h(t) = -16t² + 40t + h₀
We also use 0 for h₀, as the rocket is launched from the ground: h(t) = -16t² + 40t + 0
To find the maximum, we find the coordinates of the vertex. To do this, we first find the axis of symmetry:
-b/2a = -40/(2*-16) = -40/-32 = 1.25
This is the x-coordinate of the vertex, which represents the amount of time in seconds. We substitute this back into our function: h(1.25) = -16(1.25²) + 40(1.25) = 25