Answer to Question #1234 in Mechanics | Relativity for sabeen
Your friend has climbed a tree to a height of 6.00 m. You throw a ball vertically up to her and it is traveling at 5.00 m/s when it reaches her. What was the speed of the ball when it left your hand if you released it at a height of 1.10 m?
According to the equations of the motion under the gravitational acceleration: v=v0 - gt; h = h0 +v0t -gt2/2; for our case v=5m/s, h = 6m, h0 = 1.1m, g= 10m/s2. Thus t = (v0-5)/10= 0.1v0 - 0.5. (h-h0) = (v02-v2)/2g; 20(6-1.1) = v02 - 25; v02 = 123 v0= 11.1 m/s.
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