Answer to Question #1234 in Mechanics | Relativity for sabeen

Question

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?

Answer 1

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.