A diver jumps from a height of 24.0m into a swimming pool. What's her speed when she enters the water? (21.7m/s)
if the diver stops after sinking 3m into the water, what's the acceleration in the water? (78.4m/s^2)
vf^2=2gh=2*9.8*24
v=21.6m/s
in the water
vf^2=Vi^2+2ad
a= 21.7^2/6
thanks so much!
To calculate the speed of the diver when she enters the water, we can use the principle of conservation of energy. The initial potential energy of the diver at a height of 24.0m is converted into kinetic energy as she falls, and this kinetic energy can be used to calculate her speed.
The potential energy (PE) of an object at a certain height is given by the equation PE = mgh, where m is the mass of the object, g is the acceleration due to gravity (approximately 9.8m/s^2 on Earth), and h is the height.
In this case, we don't know the mass of the diver, but it cancels out in the calculation of speed. So we can ignore it for now.
The potential energy of the diver at a height of 24.0m is PE = mgh = 0.
The kinetic energy (KE) of an object is given by the equation KE = 0.5mv^2, where m is the mass of the object and v is its velocity.
At the point just before entering the water, the potential energy is converted into kinetic energy. So we can equate the potential energy to the kinetic energy:
mgh = 0.5mv^2
We can cancel out the mass on both sides of the equation:
gh = 0.5v^2
Rearranging the equation to solve for v:
v^2 = 2gh
Taking the square root of both sides:
v = √(2gh)
Substituting the given values:
v = √(2*9.8m/s^2 *24.0m) ≈ 21.7m/s
Therefore, the speed of the diver when she enters the water is approximately 21.7m/s.
To calculate the acceleration in the water, we can use the equation of motion:
v^2 = u^2 + 2as
where v is the final velocity (0 m/s, as the diver stops after sinking), u is the initial velocity (the speed of the diver when she enters the water, approximately 21.7 m/s), a is the acceleration in the water (what we want to find), and s is the displacement.
In this case, the final velocity is 0 m/s, the initial velocity is approximately 21.7 m/s, and the displacement is 3 m (the distance the diver sinks into the water).
Plugging these values into the equation:
0^2 = (21.7 m/s)^2 + 2a(3 m)
0 = 470.89 m^2/s^2 + 6a m
Rearranging the equation to solve for a:
6a m = -470.89 m^2/s^2
a = -78.4 m/s^2
Therefore, the acceleration in the water is approximately -78.4 m/s^2 (negative because it acts in the opposite direction to the motion).