Instantaneous rate of change at a vertex using limits ?

Question

 ball is thrown in the air. Its height from the group in meters after t seconds is modeled by h(t) = -5t^2 + 20t +1. What is the instantaneous velocity of the ball at t=2 seconds?

Answer 1

The position function of the ball is  h(t) = -5t² + 20t +1 .

The velocity is the rate change in the position function .

So now derivative the position function .

Velocity function v(t) = dh / dt .

v(t) = d/dt ( -5t² + 20t +1 )

v(t) = -10t + 20 .                   [ since the derivative of  xⁿ = n x^n-1  ] 

So the velocity function at time t is v(t) = -10t + 20 .

the velocity function at time 2 is 

                                    v(2) = -10(2) + 20 

                                    v(2) = -20 + 20 

                                    v(2) = 0 m/s .

[ v(2) = 0 m/s means that the object reaches its maximum height . ]

So the instantaneous velocity at time t = 2 sec is v(2) = 0 m/s .