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The height s of a ball in feet thrwn with an initial velocity of 80 feet er second from an initial height of 6 feet is givenas a function of the time t in seconds by
s(t)=16t^2+80t+6
(a) Graph
(b)Determine the time at which the height is maximum
(c)What is the maximum height?
Would I put this equation into my TI-86 and interpret the graph? Or am I totally wrong?
that would be a good starting point.Quote:
Originally Posted by Kayria
Alright so I did that.
And my graph is a messed up line on (5,0) Then goes up slightly to the left and down slightly to the right.
Since it only intercects the graph at that poing would that be the max?
My teacher gave us this assignment without teaching it to us..
I'm assuming this is a calculus class correct? if so, no real need to use a calculator, except to check your work. If so,
part B: take first derivative of your function, set it equal to 0 and solve for t
part C: put the answer u got in part B into your original equation to find the max height
btw ... your position function should be
the graph should be an inverted parabola.
what math class is this for? you can find this solution various ways. as for the graphing calc, it is a very useful tool. graphing it on your calculator and copying it onto your homework is one way to complete A. this graph will also show you when the ball is at its max height. just be careful in case you cant use calculators on your test!
Its for Pre-Calc. And we are urged to and can use our calcs on our testsQuote:
Originally Posted by Dubulus
in the back of the book the answer is 2.5 seconds and 106 feet.. im just not getting the parabola.
Any reasons you can think why?
i plugged it into my ti 84 and i got the correct result what are you entering in?