A diver springs upward from a board that is 4.40 m above the water. At the instant she contacts the water her speed is 14.0 m/s and her body makes an angle of 68.8 ° with respect to the horizontal surface of the water. Determine her initial velocity, both (a) magnitude and (b) direction.
To determine the initial velocity of the diver, we can use the following equations:
1. The equation for vertical displacement (Δy) is given by:
Δy = V₀y * t + (1/2) * g * t²
Where:
Δy = vertical displacement (4.40 m)
V₀y = initial vertical velocity (unknown)
t = time of flight (unknown)
g = acceleration due to gravity (9.8 m/s²)
2. The equation for horizontal displacement (Δx) is given by:
Δx = V₀x * t
Where:
Δx = horizontal displacement (0 m, as the diver returns to the same spot)
V₀x = initial horizontal velocity (unknown)
t = time of flight (unknown)
3. The equation for initial velocity (V₀) is given by:
V₀ = sqrt(V₀x² + V₀y²)
Where:
V₀ = initial velocity (unknown)
V₀x = initial horizontal velocity (unknown)
V₀y = initial vertical velocity (unknown)
Now, let's solve for the unknowns.
Step 1: Solve for time of flight (t)
Δy = V₀y * t + (1/2) * g * t²
Since the diver springs upward and ends at the same height, Δy = 0.
0 = V₀y * t + (1/2) * g * t²
We can use the quadratic formula to solve for t:
0 = (1/2) * g * t² + V₀y * t
t = [ -V₀y ± sqrt((V₀y)² - 4 * (1/2) * g * 0) ] / (2 * (1/2) * g)
t = [ -V₀y ± sqrt((V₀y)²) ] / g
t = -V₀y / g OR t = 0
Since the time cannot be negative and the diver is at a height above the water, t ≠ 0.
Therefore, t = -V₀y / g
Step 2: Solve for initial horizontal velocity (V₀x)
Since the horizontal displacement, Δx, is equal to 0, we can say that:
Δx = V₀x * t
0 = V₀x * t
Since t ≠ 0 (as we determined earlier), the initial horizontal velocity, V₀x, must also be equal to 0, as there is no horizontal motion.
Step 3: Solve for initial vertical velocity (V₀y)
We can use the given information to determine the initial vertical velocity:
V₀y = V₀ * sin(θ)
Where:
V₀y = initial vertical velocity
V₀ = initial velocity (unknown)
θ = angle of the body with respect to the horizontal surface (68.8 °)
Substituting the known values:
14.0 m/s = V₀ * sin(68.8 °)
Now, let's solve for V₀.
V₀ = 14.0 m/s ÷ sin(68.8 °)
V₀ ≈ 15.53 m/s
(a) The magnitude of her initial velocity is approximately 15.53 m/s.
(b) The direction can be determined from the angle θ. The diver springs upward, so the initial velocity is directed vertically upward at an angle of 68.8 ° above the horizontal surface of the water.
To determine the diver's initial velocity, we need to break it down into horizontal and vertical components. Let's start with the vertical component.
Given:
Height of the diving board (h) = 4.40 m
Speed (v) = 14.0 m/s
Angle (θ) = 68.8°
(a) To find the initial vertical velocity (vy), we can use the following equation:
vy = v * sin(θ)
vy = 14.0 * sin(68.8°)
vy = 14.0 * 0.9271
vy ≈ 12.98 m/s
So, the initial vertical velocity is approximately 12.98 m/s.
Now, let's calculate the horizontal component of the initial velocity.
(b) To find the initial horizontal velocity (vx), we can use the following equation:
vx = v * cos(θ)
vx = 14.0 * cos(68.8°)
vx = 14.0 * 0.3746
vx ≈ 5.246 m/s
So, the initial horizontal velocity is approximately 5.246 m/s.
Now that we have both the vertical and horizontal components of the initial velocity, let's find the magnitude and direction.
To find the magnitude of the initial velocity (v0), we can use the Pythagorean theorem:
v0 = √(vx^2 + vy^2)
v0 = √((5.246)^2 + (12.98)^2)
v0 = √(27.5534 + 168.8804)
v0 ≈ √196.4338
v0 ≈ 14.01 m/s (rounded to two decimal places)
So, the magnitude of the initial velocity is approximately 14.01 m/s.
To find the direction of the initial velocity, we can use trigonometry:
tan(θ) = vy / vx
tan(θ) = 12.98 / 5.246
tan(θ) ≈ 2.4734
Now, find the angle (α) using the inverse tangent (tan⁻¹) function:
α = tan⁻¹(2.4734)
α ≈ 67.36°
So, the direction of the initial velocity is approximately 67.36° upward from the horizontal surface of the water.
Therefore, the initial velocity of the diver is approximately (14.01 m/s, 67.36°)