Range of a projectile motion is given by
R = v cos θ / g (v sin θ + sqrt(v^2 sin^2 θ + 2gy_0)); where R = 188m, θ = 41°, g = 9.8m/s^2, y_0 = 0.9
188 = v cos 41° / 9.8 (v sin 41° + sqrt(v^2 sin^2 41° + 2 x 9.8 x 0.9)) = 0.07701(0.6561v + sqrt(0.4304 v^2 + 17.64)) = 0.05053v + 0.07701sqrt(0.4304v^2 + 17.64)
0.07701sqrt(0.4304v^2 + 17.64) = 188 - 0.05053v
0.005931(0.4304v^2 + 17.64) = 35344 - 19v + 0.002553v^2
0.002553v^2 + 0.1046 = 35344 - 19v + 0.002553v^2
19v = 35344 - 0.1046 = 35343.8954
v = 35343.8954/19 = 1860 m/s
It doesn't because when u threw it the first time, u notice that the ball eventually came to a stop because of the force that was acting upon it. Although when u throw it harder it will start out faster than the first time u threw it because u put more kinetic energy onto the ball. But the same thing happens with this ball that happened to the second ball, they both have a type of force acting upon them.
The same bird on the tree has more gravitational potential energy. This is because it is at a higher distance from the ground as it is on the tree, than when it is on the ground.
Considering also the formula for Gravitational Potential Energy GPE = mgh
For the bird on the ground, h =0, therefore GPE = m*9.8*0 = 0
For that on the tree = mgh = m*9.8*h
Of course the one on the tree has a value greater than zero.
Electricity can travel in a closed circuit.