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Expert: - 4/18/2009


Question
If a ball is thrown straight up into the air with an initial velocity of 50ft/s its height in feet after t second is given by y=50t-16t^2 Find the average velocity for the time period beginning when t=2 and lasting.

a) .5 seconds =
b) .1 seconds=
c) .01 seconds =
d) Finally based on the above results, guess what the instantaneous velocity of the ball is when t=2

I am using the formula but nothing cancels and it gets really confusing for me and I really  need help at least solving for one answer so i have a background to follow for the rest. PLEASE HELP! i'm desperate.

Answer
As can be found in http://en.wikipedia.org/wiki/Gravity
the force of gravity is 32.2.  That gives as -32.2t.
The throw gives the velocity a positive speed of 50.

The equation for speed is the 50 - 32.2t.

For (a), (b), and (c), put into the equation.
a) At t = 0.5, the velocity is 50 - 32.2(0.5) = 50 - 16.1 = 33.9.
b) At t = 0.1, the velocity is 50 - 32.2(0.1) = 50 - 3.22 = 46.78
c) At t = 0.01, the velocity is 50 - 32.2(0.01) = 50 - 3.22 = 49.678.

At t=2, I'm not sure how the above results help,
but the height is the integral of the velocity.

That is, 50t - 16.1tē.  Putting 2 in gives 50(2) - 16.1(4)
= 100 - 64.4 = 35.6.

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