Question #1234

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?

Expert's answer

According to the equations of the motion under the gravitational acceleration:

v=v_{0} - gt;

h = h_{0} +v_{0}t -gt^{2}/2;

for our case v=5m/s, h = 6m, h_{0} = 1.1m, g= 10m/s^{2}.

Thus

t = (v_{0}-5)/10= 0.1v_{0} - 0.5.

(h-h_{0}) = (v_{0}^{2}-v^{2})/2g;

20(6-1.1) = v_{0}^{2} - 25;

v_{0}^{2} = 123

v_{0}= 11.1 m/s.

v=v

h = h

for our case v=5m/s, h = 6m, h

Thus

t = (v

(h-h

20(6-1.1) = v

v

v

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