A diver springs upward with an initial speed of 2.23 m/s from a 3.0-m board. (a) Find the velocity with which he strikes the water. [Hint: When the diver reaches the water, his displacement is y = -3.0 m (measured from the board), assuming that the downward direction is chosen as the negative direction.] (b) What is the highest point he reaches above the water?
7.54
To find the velocity with which the diver strikes the water, we can use the kinematic equation for vertical motion:
vf^2 = vi^2 + 2aΔy
where:
- vf is the final velocity
- vi is the initial velocity
- a is the acceleration due to gravity (-9.8 m/s^2)
- Δy is the displacement in the y direction
(a) Let's calculate the velocity with which the diver strikes the water:
Given:
vi = 2.23 m/s (upward velocity)
Δy = -3.0 m (negative sign is used to indicate the downward displacement)
Plugging these values into the equation, we have:
vf^2 = (2.23 m/s)^2 + 2(-9.8 m/s^2)(-3.0 m)
Simplifying the equation, we get:
vf^2 = 4.9729 m^2/s^2 + 58.8 m^2/s^2
vf^2 = 63.7729 m^2/s^2
Taking the square root of both sides, we find:
vf ≈ 7.99 m/s (rounded to two decimal places)
Therefore, the velocity with which the diver strikes the water is approximately 7.99 m/s.
(b) To find the highest point the diver reaches above the water, we need to find the maximum height (h) first. The maximum height can be found using the equation:
vf^2 = vi^2 + 2aΔy
In this case, the initial velocity is the same as before (2.23 m/s), and the displacement is equal to the maximum height (h). The final velocity at the highest point is 0 m/s since the diver momentarily stops.
Plugging in the values, we have:
0^2 = (2.23 m/s)^2 + 2(-9.8 m/s^2)h
0 = 4.9729 m^2/s^2 - 19.6 h
19.6 h = 4.9729 m^2/s^2
h = 4.9729 m^2/s^2 / 19.6 m/s^2
h ≈ 0.254 m (rounded to three decimal places)
Therefore, the highest point the diver reaches above the water is approximately 0.254 m.