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?
Assume the total energy (kinetic and gravitational potential) is the same at both heights.
M*g *1.1 + (M/2)V1^2 = M*g*6.00 + (M/2)V2^2
g*(6 - 1.1) = (1/2)(5^2 -V2^2)
Solve for the velocity at 6 meter height, which I have called V2.
is g in this equation 9.8 or -9.8?
To solve this problem, we can use the principles of conservation of energy.
The initial energy of the ball is equal to the sum of its kinetic energy and potential energy. Initially, the ball only has potential energy, given by the formula PE=mgh, where m is the mass of the ball, g is the acceleration due to gravity, and h is the difference in height between the release point and the final point.
When the ball reaches your friend at a height of 6.00 m, its energy is now equal to the sum of its kinetic energy and potential energy at that point. At this point, the ball has both kinetic and potential energy given by KE=0.5mv^2 and PE=mgh respectively.
Using the principle of conservation of energy, we can equate the initial energy to the final energy:
Potential energy initially = Kinetic energy finally + Potential energy finally
So, mgh_initial = 0.5mv^2 + mgh_final
We can cancel out the mass (m) from both sides of the equation, as it appears on both sides:
gh_initial = 0.5v^2 + gh_final
Now we can solve for the velocity (v) when it left your hand. Rearranging the equation gives us:
v^2 = 2gh_initial - 2gh_final
Substituting the given values:
v^2 = 2 * 9.8 m/s^2 * (6.00 m - 1.10 m)
v^2 = 2 * 9.8 m/s^2 * 4.90 m
v^2 = 96.04 m^2/s^2
Taking the square root of both sides, we find:
v = 9.80 m/s
Therefore, the speed of the ball when it left your hand was 9.80 m/s.