A 1.13 kg hollow ball with a radius of 0.139 m, filled with air, is released from rest at the bottom of a 1.86 m deep pool of water. How high above the water does the ball shoot upward? Neglect all frictional effects, and neglect the ball's motion when it is only partially submerged.
never mind. i got it.
u would find the acceleration to find velocity of the ball coming out of the water and then, use kinematics to find the height.
I'm kind of confused about this question... so to figure out the accleration you would do F= Fb -w where Fb is the buoyancy force and F is the net force. THen you would you F=ma to find the accleration...but after that I'm kind of lost...helpp pleaseee..
To find the height above the water at which the ball shoots upward, we can use the principle of conservation of energy. We'll consider the potential energy and kinetic energy of the ball at different stages of its motion.
Here are the steps to solve this problem:
1. Determine the change in potential energy when the ball is submerged in water:
The potential energy depends on the height of the center of mass of the object above the reference level, which is the top of the water in this case. The change in potential energy (ΔPE_water) is given by the formula ΔPE_water = m * g * h, where m is the mass of the ball, g is the acceleration due to gravity (9.8 m/s²), and h is the depth of the pool (1.86 m).
ΔPE_water = (1.13 kg) * (9.8 m/s²) * (1.86 m)
2. Calculate the potential energy of the ball at the bottom of the pool:
The potential energy at the bottom of the pool is equal to the potential energy when the ball shoots upward. Hence, at this point, the ball's potential energy is maximum, and its kinetic energy is zero.
Potential energy at the bottom (PE_initial) = m * g * 2r, where r is the radius of the ball.
PE_initial = (1.13 kg) * (9.8 m/s²) * (2 * 0.139 m)
3. Find the height above the water at which the ball shoots upward:
At the maximum height, the potential energy will be zero, and the kinetic energy will be maximum. This occurs when the ball reaches its highest point in the motion.
Potential energy at the highest point (PE_final) = 0
Kinetic energy at the highest point (KE_final) = (1/2) * m * v², where v is the velocity of the ball at the highest point.
Since the potential energy is equal to the kinetic energy at the highest point, we can write:
PE_final = KE_final
0 = (1/2) * (1.13 kg) * v²
Solving for v, we get:
v = √0 / (1/2 * 1.13 kg)
v = 0 m/s
So, the velocity at the highest point is zero.
4. Determine the height above the water at which the ball shoots upward:
The height above the water at which the ball shoots upward is given by the following formula:
h_above_water = (PE_initial - PE_final) / (m * g)
h_above_water = (PE_initial - 0) / (1.13 kg * 9.8 m/s²)
Finally, plug in the values and calculate:
h_above_water = (((1.13 kg) * (9.8 m/s²) * (2 * 0.139 m)) - 0) / ((1.13 kg) * (9.8 m/s²))
The value obtained will give the height above the water at which the ball shoots upward.