A diver jumps off of a diving board. She leaves the board with a velocity of 3.70 m/sec in the upward direction. How far will she rise above the diving board before she reverses her motion and starts to fall downward?
PE at top= KE at start
mgh=1/2 m v^2
solve for h.
To find the distance the diver will rise above the diving board before reversing her motion, we first need to calculate the maximum height she reaches.
We can use the kinematic equation for the motion in the vertical direction:
vf^2 = vi^2 + 2ad
Where:
- vf is the final velocity (which is zero when she reaches the peak)
- vi is the initial velocity (3.70 m/s upward)
- a is the acceleration due to gravity (-9.8 m/s^2, since she is moving against the force of gravity)
- d is the displacement or the height we are trying to find
Rearranging the equation, we have:
0 = (3.70 m/s)^2 + 2(-9.8 m/s^2)d
Simplifying the equation:
0 = 13.69 m^2/s^2 - 19.6 m/s^2 * d
Solving for d:
19.6 m/s^2 * d = 13.69 m^2/s^2
d = 13.69 m^2/s^2 / 19.6 m/s^2
d ≈ 0.698 meters
Therefore, the diver will rise approximately 0.698 meters above the diving board before she reverses her motion and starts to fall downward.