If a ball is thrown into the air with a velocity of 41 ft/s, its height (in feet) after t seconds is given by y = 41t − 16t2. Find the velocity when t = 1.

Answer 1

y = 41t - 16 t^2
differentiating with respect to t,
dy/dt = 41 - 32 t
dy/dt is the velocity of the ball
at t = 1,
dy/dt = 41 - 32 = 9
so the velocity is 9 ft/s (upwards)