A projectile is fired straight upward from the ground. Its height h in feet after t seconds is given by the equation h(t)=-16t^2+96t. Find the maximum height of the projectile. I think i know how to do it but i am not sure if i did it right.
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Take the derivative of the equation and set it equal to zero. This will tell you when the slope of the line is zero (the vertex).
You should get -32x + 96= 0 for the derivative. Solve for x and you get x=3
Plug this back into your original equation to find the y value (the height)
-16(3)^2 + 96(3)= 144
So the maximum height is 144 ft.
You should get -32x + 96= 0 for the derivative. Solve for x and you get x=3
Plug this back into your original equation to find the y value (the height)
-16(3)^2 + 96(3)= 144
So the maximum height is 144 ft.