# Suppose a rock is thrown straight up in the air with initial velocity of 40ft/sec. It's height after t seconds is given by y=40t-16t^2. Find the velocity when t=2 and at what time will the rock hit the surface? With what velocity will it hit the surface?

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.

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