velocity prob?

Question

A rock is thrown straight up with an intial velocity of 19.6 m/s. What time interval elapses between the rock's being thrown and its return to the original launch point?

please help! thanks!!

Answer 1

pretty obviously, they want to use 9.8 m/s^2 as gravity, which is fine.

distance travelled in physics is:
s(t) = v0*t + 1/2*g*t*t
Now, we want total distance travelled to equal 0 (right back where we started), so we plug in:

0 = 19.6*t + 1/2 * (-9.8) * t^2 (note gravity is NEGATIVE if initial velocity is positive, they're in opposite directions).
-4.9t^2 + 19.6t = 0
t * (-4.9t + 19.6 ) = 0, so this is true when t = 0 (at the time of throwing the rock) and when (-4.9t + 19.6 = 0)
-4.9t + 19.6 = 0
4.9t = 19.6
t = 4 seconds. 4 seconds after the rock is thrown, it's back at release point.

Answer 2

question a million C = 2 ? r C = 24 ? cm a million rev in a million hour----->24 revs in a million day distance = 24? x 24 cm distance = 1810 cm (to nearest entire quantity) question 2 v = 16 km / h v = 16,000 / 3,six hundred m/s v = one hundred and sixty / 36 m/s v = ? r ? = v / r ? = (one hundred and sixty / 36 ) / 30 radians / sec ? = 16 / 108 radians / sec ? = 0.148 radians / sec