Physics t = sqrt 2d/g?
ok so my physics teacher was talking about the formula and i sorta... dosed off ( im a baaaad student =( . can't pay attention ) anyways i wuz writing down notes and he gave this example
sqrt 2*30 / 980 cm/s = 2(.255) = .510 s
ahhhh, any physics buff students out that can help me ? im stuck because i have 30 for d , bu i don't know how he got from sqrt 2*30 / 980 cm/s to 2(.255)
i know its my fault for not being able to pay attention, so if you can help i'd be reaaally really grateful. thakn you ~~~!!!! ^-^
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Man, you could give Mr. Vick lessons in being humble. Anyway, the equation starts with
d = 1/2 g*t, where g is the accleration of gravity of 980 cm/SEC^2, t is time and d= distance from a still "drop"(an object is released with no initial velocity). From that you can get to t= sqrt (2*d/g) and do the numerics. You are solving for the time for the object to drop 30 cm.