h(t)=-16t^2+576t. After how many seconds does the projectile reach its maximum height?

Question

A projectile is thrown upward so that its distance is above the ground after t seconds is given by the function h(t)=-16t^2+576t. After how many seconds does the projectile reach its maximum height?

Answer 1

The function is h =-16t²+576t, where h = height and t =time.

Derivative respect to 't' to each side.

dh/dt =d/dt(-16t²+576t)

         =-16 d/dt(t²)+576d/dt(t)

         =-16(2t)+576(1)

         =-32t+576     ....... (1)

To find the critical or key numbers, to make the first derivative equal to zero or dh/dt = 0.

-32t+576 = 0

32t= 576

t = 18 s

The critical numbers is t = 18 s

Derivative respect to 't' to each side of the equation (1)

d²h/dt² = -32<0

             = Local maximum.

The time for maximum height at t = 18s.