A diver springs upward from a board that is three meters above the water. At the instant she contacts the water her speed is 8.80 m/s and her body makes an angle of 76.0° with respect to the horizontal surface of the water. Determine her initial velocity, both magnitude and direction.
m/s (magnitude)
° (direction)
Well, it seems like the diver really knows how to make a splash! Let's dive into the problem (pun intended).
To determine the diver's initial velocity, we need to break down the given information. We know her final speed, the angle of her body with respect to the horizontal surface, and the height from where she springs.
First, let's find the initial vertical velocity (Viy). We can use the kinematic equation:
Vf^2 = Vi^2 + 2ad
Where Vf is the final velocity (8.80 m/s), Vi is the initial velocity (which we're trying to find), a is the acceleration due to gravity (-9.8 m/s^2), and d is the distance (3 meters).
Plugging in the values, we get:
(8.80 m/s)^2 = Vi^2 + 2(-9.8 m/s^2)(-3 meters)
Simplifying this equation, we find:
Vi^2 = 68.48 m^2/s^2 + 58.8 m^2/s^2
Vi^2 = 127.28 m^2/s^2
Now, let's find the initial horizontal velocity (Vix). Since there is no horizontal acceleration, the initial and final horizontal velocities are the same. Thus:
Vix = Vfx
Vi * cos(theta) = Vf
Vi * cos(76.0°) = 8.80 m/s
Now, we can solve for Vi:
Vi = 8.80 m/s / cos(76.0°)
Calculating this, we find:
Vi ≈ 32.86 m/s
So, the magnitude of the diver's initial velocity is approximately 32.86 m/s.
Now, let's talk about the direction. The angle we found, 76.0°, represents the angle of the diver's body with respect to the horizontal surface of the water. Since angles are measured counterclockwise from the positive x-axis, we can say the direction (angle) is 76.0°.
Therefore, her initial velocity is approximately 32.86 m/s in a direction of 76.0°.
Now go make a splash with this knowledge!
To determine the initial velocity of the diver, we can use the concept of projectile motion.
Let's break down the given information:
- The diver springs upward from a board that is 3 meters above the water.
- The speed of the diver at the instant she contacts the water is 8.80 m/s.
- The angle her body makes with respect to the horizontal surface of the water is 76.0°.
We can analyze the motion in the vertical and horizontal directions separately.
Vertical Motion:
The vertical motion of the diver can be treated as free fall, considering only the acceleration due to gravity. In this case, we can use the kinematic equation:
vf = vi + gt
where:
- vf is the final velocity in the vertical direction (which is 0 m/s when the diver contacts the water),
- vi is the initial velocity in the vertical direction,
- g is the acceleration due to gravity (approximately 9.8 m/s^2),
- t is the time taken for the diver to reach the water.
Since the final velocity is 0 m/s, we can rearrange the equation to solve for the initial velocity:
vi = -gt
Substituting the given values:
vi = -(9.8 m/s^2) * t
Horizontal Motion:
The horizontal motion of the diver is constant and unaffected by gravity. Therefore, the horizontal velocity remains constant throughout the motion, and the initial horizontal velocity (vix) remains the same as the final horizontal velocity (vfx).
The magnitude of the initial velocity (vi) can now be found using trigonometry. By breaking down the initial velocity into its horizontal and vertical components, we can use the relationships between the sides and angles of a right triangle.
Considering the angle of 76.0°, the vertical component of the velocity (viy) can be calculated using:
viy = vi * sin(theta)
And the horizontal component of the velocity (vix) can be calculated using:
vix = vi * cos(theta)
To find the magnitude of the initial velocity (vi), we can use the Pythagorean theorem:
vi = sqrt(vix^2 + viy^2)
Finally, to determine the direction of the initial velocity, we can use inverse trigonometric functions:
direction = arctan(viy/vix)
Now let's calculate the values step by step:
1. Calculate the time taken (t) using the vertical motion equation:
t = vf / g
= 0 m/s / (9.8 m/s^2)
= 0 s (as the numerator is zero)
2. Calculate the vertical component of the velocity (viy):
viy = vi * sin(theta)
= -(9.8 m/s^2) * t * sin(76.0°)
3. Calculate the horizontal component of the velocity (vix):
vix = vi * cos(theta)
= -(9.8 m/s^2) * t * cos(76.0°)
4. Calculate the magnitude of the initial velocity:
vi = sqrt(vix^2 + viy^2)
5. Determine the direction of the initial velocity:
direction = arctan(viy/vix)
Now you can substitute the values into the equations and calculate the final result.