Physics t = sqrt 2d/g?
Hi everyone, I need some help with a physics formula my teacher explained in class. Unfortunately, I kind of zoned out (I know, bad habit!), so I'm a little confused about how it works.
The example he gave involved the formula ( t = \sqrt{\frac{2d}{g}} ), and here’s what I managed to write down:
[ t = \sqrt{\frac{2 \cdot 30}{980 \, \text{cm/s}^2}} = 2(0.255) = 0.510 ]
I understand that ( d = 30 ), but I’m stuck on how he got from ( \sqrt{\frac{2 \cdot 30}{980}} ) to ( 2(0.255) ). I feel like I missed an important step or explanation.
If anyone could break this down for me or point out what I’m missing, I’d really appreciate it! Thank you so much in advance! 😊
1 Answers
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.
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