A diver springs upward from a board that is 2.98 meters above the water. At the instant she contacts the water her speed is 9.39 m/s and her body makes an angle of 70.4° with respect to the horizontal surface of the water. Determine her initial velocity.
Her vertical velocity component when she hits the water is
Vy' = 9.39 sin70.4
Her horizontal velocity component is
Vx = 9.39 cos70.4
The latter component remains constant during the dive.
For the initial vertical velocity component, use
Vyo^2/2 + g*2.98 m = Vy'^2/2
Use the initial horizontal and vertical velocity components, Vx and Vyo, ffor the initial velocity.
To determine the diver's initial velocity, we can use the principles of projectile motion. We know the height of the board (2.98 meters) and the speed at which the diver enters the water (9.39 m/s) at an angle of 70.4° with respect to the horizontal surface of the water.
The initial velocity of the diver can be broken down into its horizontal and vertical components. Let's first determine the vertical component of the initial velocity.
The vertical motion of the diver can be described using the equation:
h = ut + (1/2)gt^2
Where:
h = height (2.98 meters)
u = initial vertical velocity (unknown)
t = time of flight (unknown)
g = acceleration due to gravity (-9.8 m/s^2)
Since the diver springs upward and then falls downward, the total time of flight would be twice the time it takes to reach the maximum height.
We can find the time of flight using the formula:
t = (2u * sinθ) / g
Where:
θ = angle with respect to the horizontal surface (70.4°)
g = acceleration due to gravity (-9.8 m/s^2)
Substituting this value of time into the previous equation, we can solve for the initial vertical velocity:
h = u * t + (1/2)g * t^2
Rearranging the equation and substituting the known values:
2.98 meters = (u * t) + (1/2)(-9.8 m/s^2)(t^2)
Simplifying the equation:
4.9t^2 + ut - 2.98 = 0
This is a quadratic equation in terms of 't'. Solving this equation will give us the time of flight, 't'.
Once we have the time of flight, we can use it to find the initial vertical velocity using the formula derived earlier:
u = (h - (1/2)gt^2) / t
Substituting the known values, we can find the initial vertical velocity:
u = (2.98 meters - (1/2)(-9.8 m/s^2)(t^2)) / t
Now that we have the vertical component of the initial velocity, we can find the horizontal component.
The horizontal component of velocity remains constant throughout the motion. We can use the formula:
v = u*cosθ
Where:
v = horizontal component of velocity (unknown)
u = initial velocity (vertical component)
θ = angle with respect to the horizontal surface (70.4°)
Solving for the horizontal component of velocity:
v = u * cosθ
Finally, we can find the magnitude of the initial velocity using the horizontal and vertical components:
Initial velocity (V) = √(v^2 + u^2)
Substituting the known values, we can find the initial velocity.