# What is the instantaneous velocity of the ball at t=2 seconds if a ball is thrown in the air and its height from the ground in meters after t seconds is modeled by #h(t)=-5t2+20t+1#?

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To find the instantaneous velocity of the ball at ( t = 2 ) seconds, we need to find the derivative of the height function ( h(t) ) with respect to time ( t ), which gives us the velocity function. Then, we can evaluate this velocity function at ( t = 2 ) seconds to find the instantaneous velocity.

The height function is given by ( h(t) = -5t^2 + 20t + 1 ).

The velocity function, which is the derivative of the height function, is given by ( v(t) = h'(t) ).

Taking the derivative of ( h(t) ) with respect to ( t ), we get:

[ h'(t) = \frac{dh}{dt} = -10t + 20 ]

Now, we can evaluate ( v(t) ) at ( t = 2 ) seconds:

[ v(2) = -10(2) + 20 ] [ v(2) = -20 + 20 ] [ v(2) = 0 ]

So, the instantaneous velocity of the ball at ( t = 2 ) seconds is ( 0 ) meters per second.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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