A diver springs upward from a board that is three meters above the water. At the instant she contacts the water her speed is 9.6 m/s and her body makes an angle of 84 degrees with respect to the horizontal surface of the water .
I'd score it 8 out of 10, but don't know the degree of difficulty.
By the way, what is your question?
In a two-dimensional tug-of-war, Alex, Betty, and Charles pull horizontally on an automobile tire at the angles shown in the picture. The tire remains stationary in spite of the three pulls. Alex pulls with force of magnitude 213 N, and Charles pulls with force of magnitude 178 N. Note that the direction of is not given. What is the magnitude of Betty's force if Charles pulls in (a) the direction drawn in the picture or (b) the other possible direction for equilibrium?
you do the 9.6m/s^2 - (2(-9.8)(-3)
the -3 because the board is 3 meters above the water so the y change is 3 down so that's -3
(92.16)-(58.8)
sqrt(33.36)
answer: 5.78m/s
To solve this problem, we can use the principles of projectile motion.
Let's break down the given information:
Initial height above the water (h): 3 meters
Initial speed (v0): 9.6 m/s
Angle of the diver's body with respect to the horizontal (theta): 84 degrees
We need to find the horizontal and vertical components of the diver's velocity just before contacting the water and then use them to calculate the time of flight and range.
Step 1: Calculate the vertical velocity component (v0y) just before contacting the water.
v0y = v0 * sin(theta)
Using the given values:
v0y = 9.6 m/s * sin(84 degrees) ≈ 9.4 m/s
Step 2: Calculate the time of flight (t) from the vertical motion.
We can use the fact that the time of flight for an object in projectile motion is double the time taken to reach the maximum height.
Using the equation:
t = 2 * (v0y / g)
where g is the acceleration due to gravity (approximately 9.8 m/s^2),
t = 2 * (9.4 m/s / 9.8 m/s^2) ≈ 1.92 s
Step 3: Calculate the horizontal distance traveled (range).
To find the horizontal distance traveled by the diver, we need to calculate the horizontal velocity component (v0x).
v0x = v0 * cos(theta)
Using the given values:
v0x = 9.6 m/s * cos(84 degrees) ≈ 1.58 m/s
Using the equation:
range = v0x * t
Substituting the values:
range = 1.58 m/s * 1.92 s ≈ 3.04 meters
Therefore, the diver will land in the water approximately 3.04 meters away from the edge of the board.