An object is launched at 20 m/s from a height of 65 m. The equation for the height (h) in terms of time (t) is given by h(t) = -4.9t² + 20t + 65. What is the object's maximum height?

Question

anonymous · Answer

Ok so you have the equation of the height wich is a quadratic equation, you solved it for t, and you take the positive value (cause it's a time and a negative time have no physics meaning). If i'm not wrong the positive time is 6.21 (s)

anonymous · Answer

Now put that value of time in the equation of the height and it will give a value for heigh, you add the 65 first meters and you have the maximum height

anonymous · Answer

i think that's the solution

anonymous · Answer

ok, just one mistake, you don't have to add the 65 meters, cause it's the independent term of the height equation, if you notice;
h=1/2(-9.8)*t^2+v*t+h(0)