A coin is tossed upward with an initial speed of 7.5 m/s2. In the absence of air

Question

A coin is tossed upward with an initial speed of 7.5 m/s2. In the absence of air resistance, how high does the coin go above the point of release? What is the total time it is in the air before returning yo it's releasing pint?

Explanation / Answer

vf^2 - vi^2 = 2 g y ==> 0^2 - (7.5*7.5) = 2 * (-9.81) * (h) ==> h = 2.87 m -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- t = 2 * vi/g = 2*7.5/9.81 = 1.53 s

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