Posted by **kayla** on Saturday, February 9, 2013 at 7:13pm.

A ball is thrown vertically upward, which is the positive direction. A little later it returns to its point of release. The ball is in the air for a total time of 8.26 s. What is its initial velocity? Neglect air resistance.

- west chester university -
**Reiny**, Saturday, February 9, 2013 at 8:41pm
Its vertical distance is given by a parabola.

Since we know the t-intercepts are 0 and 8.26 ,and we know a =-4.9 m/sec^2

the equation is

s = -4.9t(t-8.26)

= -4.9t^2 + 40.472 , where m is in metres

v = -9.8t + 40.472

when t = 0 , v = 40.472

initial velocity is 40.72 m/sec

Wow, that is quite a toss!

BTW, if you are working in feet,

a = -32 ft/sec^2

duplicate the above steps, not much has to be changed.

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