A diver springs vertically upward with an initial speed of 1.7 m/s from a diving board
that is 3.2 m above the water. Find the speed with which she enters the water, assuming
completely vertical motion.
mv₀²/2= mgh
h= v₀²/2g
mg(H+h) = mv²/2
mgH +mv₀²/2 = mv²/2
v=sqrt{ v₀² +gH} =
= sqrt{1.7²+9.8•3.2} = 5.85 m/s
To find the speed with which the diver enters the water, we need to use the principles of kinematics and apply the laws of motion.
Step 1: Understand the problem and identify the given information:
- Initial speed of the diver (u) = 1.7 m/s (upward)
- Height of the diving board (h) = 3.2 m
- Acceleration due to gravity (g) = 9.8 m/s^2 (downward)
Step 2: Determine the time taken by the diver to reach the peak of the motion:
When the diver reaches the peak, her final velocity (v) will be zero. We can use the following equation to find the time taken (t):
v = u + gt
0 = 1.7 - 9.8t
9.8t = 1.7
t = 1.7 / 9.8 ≈ 0.1735 s
Step 3: Calculate the time taken for the diver to reach the water:
Since the motion is symmetric, the time taken to go up to the peak (t) will be equal to the time taken to come down. Therefore, the total time taken for the diver to reach the water is twice the time taken to reach the peak, which is 2t ≈ 2 * 0.1735 ≈ 0.347 s.
Step 4: Find the final velocity when the diver hits the water:
We can use the following equation to find the final velocity (v) of the diver when she enters the water:
v = u + gt
v = 1.7 + 9.8 * 0.347
v = 1.7 + 3.4026
v ≈ 5.10 m/s
Therefore, the speed with which the diver enters the water, assuming completely vertical motion, is approximately 5.10 m/s.