example a ball is thrown straight up with velocity given by t 32t 20fts where t is measured in s

The solution for the problem above problem is the initial velocity is equal to 42 feet per secon

So here we're told that a ball is thrown with an initial height of 510 feet and a velocity of 16

position function. So this is going to represent the position and we know a couple pieces of inf

So I have a position function, negative 16t squared plus v sub 0t plus s sub 0. And I'm told a b

Alright, so we are given our position function is s of t equal to negative 16 t squared plus ini

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Example: A ball is thrown straight up with a velocity given by v(t) = 32t + 20 ft/s, where t is measured in seconds. Find and interpret âˆ«v(t) dt. Explanation: Using the Second Fundamental Theorem of Calculus, we have âˆ«(32t + 20) dt = 16t^2 + 20t + C. Thus, if a ball is thrown straight up into the air with velocity v(t), the height of the ball, 1 second later, will be 4 feet above the initial height. Note that the ball has traveled much farther. It has gone up to its peak and is falling down, but the difference between its height at t=0 and t=1 is 4 ft.

          Example: A ball is thrown straight up with a velocity given by v(t) = 32t + 20 ft/s, where t is measured in seconds. Find and interpret âˆ«v(t) dt.

Explanation: Using the Second Fundamental Theorem of Calculus, we have
âˆ«(32t + 20) dt = 16t^2 + 20t + C.

Thus, if a ball is thrown straight up into the air with velocity v(t), the height of the ball, 1 second later, will be 4 feet above the initial height. Note that the ball has traveled much farther. It has gone up to its peak and is falling down, but the difference between its height at t=0 and t=1 is 4 ft.

example a ball is thrown straight up with velocity given by t 32t 20fts where t is measured in seconds find and interpret j t dt explanation using the second fundamental theorem of calculus 93547

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