a diver of mass 101 kg jumps upward off a diving board into water. Diving board is 6m above water. Driver has a speed of 1.2m/s. based on speed how high above the board will she rise?
V^2 = Vo^2 + 2g*h
V = 0
Vo = 1.2 m/s
g = -9.8 m/s^2
h = ?
To find how high above the diving board the diver will rise based on her speed, we can use the principles of conservation of energy.
First, let's calculate the initial potential energy of the diver when she jumps off the diving board. The potential energy (PE) is given by the formula PE = m * g * h, where m is the mass, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height above the reference point (determined by the diving board).
PE = 101 kg * 9.8 m/s^2 * 6 m
PE = 5928 J
Next, let's calculate the initial kinetic energy of the diver using the formula KE = (1/2) * m * v^2, where m is the mass and v is the speed.
KE = 0.5 * 101 kg * (1.2 m/s)^2
KE = 73.032 J
Initially, the total mechanical energy (potential energy plus kinetic energy) is conserved, assuming no external work done on the system. Therefore, the initial mechanical energy of the system is equal to the final mechanical energy of the system when the diver is at the highest point.
Considering that the final mechanical energy is equal to the potential energy at the highest point, we equate the two as follows:
PE = Total Mechanical Energy at highest point
5928 J = 73.032 J + PE_highest
Solving for PE_highest:
PE_highest = 5928 J - 73.032 J
PE_highest = 5854.968 J
Finally, let's calculate the height above the diving board at the highest point using the formula PE_highest = m * g * h_highest. Rearranging this equation, we find:
h_highest = PE_highest / (m * g)
h_highest = 5854.968 J / (101 kg * 9.8 m/s^2)
h_highest ≈ 5.994 m
Therefore, based on the diver's speed of 1.2 m/s, she will rise approximately 5.994 meters above the diving board.