# 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...

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?

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### 1 Answer

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.**