a ballon carrying a ball is moving vertically upward with velocity 12 m/sec. when the ballon is at height 65m the ball is dropped,with what speed ball hit the ground?
m g h + 1/2 m v^2 = 1/2 m V^2
(2 * 9.8 * 65) + 12^2 = V^2
To find the speed at which the ball hits the ground, we can use the principle of conservation of energy.
When the ball is dropped from a height, it will experience a change in potential energy, which gets converted to kinetic energy. We can equate the change in potential energy to the gained kinetic energy to find the velocity of the ball when it hits the ground.
The potential energy of an object at a height h is given by the equation:
Potential energy (PE) = m * g * h
Where m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the vertical height.
The kinetic energy of an object is given by the equation:
Kinetic energy (KE) = (1/2) * m * v^2
Where m is the mass of the object and v is the velocity of the object.
Initially, the potential energy of the ball is m * g * h and its kinetic energy is (1/2) * m * (12)^2, as it is moving upward with a velocity of 12 m/s.
When the ball hits the ground, its potential energy becomes zero. So, the remaining energy is in the form of kinetic energy, given by (1/2) * m * v^2.
Setting the initial kinetic energy equal to the final kinetic energy, we have:
(1/2) * m * (12)^2 = (1/2) * m * v^2
Simplifying the equation, we get:
144 = v^2
Taking the square root of both sides, we have:
v = √144
v = 12 m/s
Therefore, the ball will hit the ground with a speed of 12 m/s.