Answer:
The height from which the diver jumped can be calculated using the laws of motion and the principle of conservation of energy.
Let's assume that the diver jumped from a platform with an initial velocity of zero. The potential energy (PE) of the diver at the top of the platform is given by:
PE = mgh
Where m is the mass of the diver, g is the acceleration due to gravity (9.8 m/s^2), and h is the height from which the diver jumped.
When the diver jumps, the potential energy is converted into kinetic energy (KE) as the diver moves downwards. The kinetic energy of the diver just before hitting the water is given by:
KE = (1/2)mv^2
Where v is the velocity of the diver just before hitting the water.
According to the principle of conservation of energy, the potential energy at the top of the platform is equal to the kinetic energy just before hitting the water. Therefore, we can equate the two equations above and solve for h:
mgh = (1/2)mv^2
h = (1/2)v^2/g
We need to find the value of v to calculate h. We can use the kinematic equation:
v^2 = u^2 + 2as
Where u is the initial velocity (which is zero in this case), a is the acceleration due to gravity (9.8 m/s^2), and s is the distance travelled by the diver (which is equal to h).
Substituting the values, we get:
v^2 = 2gh
v^2 = 2 * 9.8 * h
v^2 = 19.6h
Therefore:
h = v^2/(19.6)
Now, let's assume that the velocity of the diver just before hitting the water was 10 m/s (a reasonable value for a diving competition). Then:
h = (10^2)/(2*9.8) = 51 meters
Therefore, the height from which the diver jumped is approximately 51 meters.