A ball of mass 0.118 kg is dropped from a height 3.2 m above the ground.
The acceleration of gravity is 9.8 m/s? Neglecting air resistance, determine the speed of the ball when it is at a height 0.799 m above the ground
Answer in units of m/s.
To determine the speed of the ball when it is at a height of 0.799 m above the ground, we can use the principle of conservation of mechanical energy.
First, let's find the potential energy of the ball when it is at a height of 3.2 m above the ground. The potential energy (PE) can be calculated using the formula:
PE = mass * gravitational acceleration * height
PE = 0.118 kg * 9.8 m/s^2 * 3.2 m
PE = 3.68896 J
Next, let's find the potential energy of the ball when it is at a height of 0.799 m above the ground. Using the same formula, we have:
PE = 0.118 kg * 9.8 m/s^2 * 0.799 m
PE = 0.92084624 J
Since the ball is dropped from rest, its initial kinetic energy (KE) will be zero. Therefore, the total mechanical energy (E) of the ball will be equal to the potential energy at any height.
E = PE
Using this principle, we can equate the potential energy at a height of 3.2 m to the potential energy at a height of 0.799 m:
3.68896 J = 0.92084624 J + KE
Simplifying the equation:
KE = 3.68896 J - 0.92084624 J
KE = 2.76811376 J
Finally, we can find the speed of the ball when it is at a height of 0.799 m using the kinetic energy formula:
KE = (1/2) * mass * velocity^2
Rearranging the equation to solve for velocity:
velocity = sqrt((2 * KE) / mass)
Substituting the values:
velocity = sqrt((2 * 2.76811376 J) / 0.118 kg)
velocity ≈ sqrt(46.4692 J/kg)
velocity ≈ 6.8184 m/s
Therefore, the speed of the ball when it is at a height of 0.799 m above the ground is approximately 6.8184 m/s.