When the rock is thrown straight up in the air with an initial velocity 40 ft/sec its height after t seconds is given by y = 40*t - 16t^2.

The velocity of the object after t seconds is given by `dy/dt = 40 - 32t`

At t = 2 seconds,...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

When the rock is thrown straight up in the air with an initial velocity 40 ft/sec its height after t seconds is given by y = 40*t - 16t^2.

The velocity of the object after t seconds is given by `dy/dt = 40 - 32t`

At t = 2 seconds, the velocity of the object is 40 - 64 = -24 ft/sec. A negative velocity indicates that the velocity is in the downwards direction.

When the object strikes the ground the value of y = 0. Solving the equation 40t - 16y^2 = 0

=> 5t - 2t^2 = 0

=> t(5 - 2t) = 0

=> t = 5/2 and t = 0

t = 0 represents the instant when the object was thrown. It rises up and strikes the ground after 2.5 seconds. The velocity of the object after 2.5 s is -40 ft/s.

**The object strikes the ground after 2.5 seconds and its velocity is 40 ft/s in the downward direction.**