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motion09

Joined: 19 Sep 2010
Posts: 4

 Posted: Tue Mar 29, 2011 12:58 am    Post subject: Kinematics Problem An astronaut on the moon throws a wrench straight up at 4.0 m/s. Three seconds later it falls downwards at a velocity of 0.8 m/s. a) What was the acceleration of the wrench after it left the astronaut's hand? b) How high above the point from which it was released was the wrench at 3.0 s? c) How long would it take the wrench to return to the position from which it was thrown? I get how to do a and b, but not c. Do we use the same variables as part a or not, since the ball is going in a different direction? Thanks in advance!
hepcj

Joined: 23 Jun 2007
Posts: 125

 Posted: Sun Apr 03, 2011 5:27 pm    Post subject: u = 4.0 m/s t = 3 s v = 0.8 m/s v = u + a t a = (v - u)/t = 3.2/3 = -1.06667 m/s^2 b) s = u t + 0.5 a t^2 s = 4.0 x 3 + 0.5 x -1.06667 x (3 x 3) = 12 - 4.8 = 7.2 m c) u =4.0 v = 0 a = -1.06 m/s^2 v = u + a t t = (v-u)/a t = 3.749 s This is the time it takes to reach its maximum vertical height. What goes up must come down so it will also take the same time to return. The total time is therefore, t = 2 x 3.749 s = 7.5 s
motion09

Joined: 19 Sep 2010
Posts: 4

 Posted: Mon Apr 04, 2011 9:53 pm    Post subject: Thanks!
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