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 − 16t^2. Find the velocity when t = 1.

Answer 1

Velocity is the derivative of position.

y = 41t -16t²

dy/dt =41 -32t

at t=1

dy/dt =41 -32(1) =9

so velocity will be 9 ft/s