If a ball is thrown into the air with a velocity of 43 ft/s, its height (in feet) after t seconds is given by y = 43t − 16t^2. Find the velocity when t = 2.

Answer 1

Hello Emanuel,

y = 43t - 16t² gives the height of the ball at the time, t, after the ball is released. This is a function about position - in this case, height.

The derivative of the position function yields the function for velocity of the object.

y' = 43 - 32t

The velocity at t = 2: 43 - 32(2) = -21 ft/ sec.

The velocity is negative because the object is slowing down from the original velocity due to the effect of gravity.

Regards,

Anita A.

Answer 2

Velocity is given by dy/dt = 43 - 32t.

Evaluate at t=2...to see that the ball is falling.