A diver runs horizontally off the end of a diving board with an initial speed of 1.75 sec. If the diving board is 3.0 m above the water line, what is the diver’s speed just before she enters the water?
7.87
To solve this problem, we can use the principles of projectile motion. We know the initial vertical velocity is 0 m/s since the diver runs horizontally off the diving board.
Step 1: Find the time it takes for the diver to hit the water.
We can use the equation of motion for vertical motion: d = v_i*t + 0.5*a*t^2.
Since the diver starts at a height of 3.0 m and falls vertically downward, the displacement (d) is -3.0 m (negative because it is downward).
The acceleration (a) due to gravity is -9.8 m/s^2 (negative because it is downward).
Plugging in these values, we get: -3.0 m = 0*t + 0.5*(-9.8 m/s^2)*t^2.
Simplifying the equation, we have: -4.9*t^2 = -3.0.
Dividing both sides by -4.9, we get: t^2 = 0.612244898.
Taking the square root of both sides, we find: t ≈ 0.78 s.
Step 2: Find the horizontal distance traveled.
Since the divergence runs horizontally, the horizontal distance traveled is the same as the distance traveled by the diver in the horizontal direction.
The horizontal distance can be found using the equation: d = v*t, where d is the horizontal distance, v is the horizontal velocity, and t is the time.
Since the diver's initial speed is 1.75 m/s, the horizontal distance can be calculated as: d = 1.75 m/s * 0.78 s = 1.365 m.
Step 3: Find the vertical velocity just before entering the water.
The vertical velocity just before entering the water can be found using the equation: v_f = v_i + a*t.
Since the initial velocity (v_i) is 0 m/s and the acceleration is -9.8 m/s^2, plugging in the values gives us:
v_f = 0 m/s + (-9.8 m/s^2) * 0.78 s = -7.644 m/s (negative because it is downward).
Finally, we can find the magnitude of the speed just before the diver enters the water.
The magnitude of the speed (v) can be calculated using the Pythagorean theorem: v = sqrt(v_horizontal^2 + v_vertical^2).
Since the horizontal velocity is 1.75 m/s and the vertical velocity is -7.644 m/s, we have:
v = sqrt((1.75 m/s)^2 + (-7.644 m/s)^2) ≈ 7.89 m/s.
Therefore, the diver's speed just before she enters the water is approximately 7.89 m/s.
To find the diver's speed just before entering the water, we need to use the principle of conservation of energy.
The initial kinetic energy of the diver is converted into potential energy and kinetic energy just before entering the water. We can assume there is no loss of energy in the process.
Let's break down the problem and calculate step by step:
1. Calculate the potential energy at the top of the diving board.
Potential energy (PE) = mass (m) * gravitational acceleration (g) * height (h)
Since we only have the height, and the question doesn't provide the mass of the diver, we can ignore the mass for now.
Gravitational acceleration (g) is approximately 9.8 m/s².
PE = m * g * h
Given:
Height (h) = 3.0 m
PE = m * 9.8 m/s² * 3.0 m
2. Calculate the initial kinetic energy of the diver.
Kinetic energy (KE) = 1/2 * mass * velocity²
We are given the initial speed, so we can calculate the initial kinetic energy.
Velocity (v) = 1.75 m/s
KE = 1/2 * m * (1.75 m/s)²
3. Set the potential energy equal to the kinetic energy just before entering the water.
PE = KE
m * g * h = 1/2 * m * v²
4. Cancel out the mass from both sides.
g * h = (1/2) * v²
5. Solve for v (speed just before entering the water).
v² = (2 * g * h)
v² = 2 * 9.8 m/s² * 3.0 m
v = √(2 * 9.8 m/s² * 3.0 m)
Calculating the value:
v = √(2 * 9.8 m²/s² * 3.0 m)
v ≈ √(58.8 m²/s²)
v ≈ 7.67 m/s
Therefore, the diver's speed just before entering the water is approximately 7.67 m/s.
1.75 sec. should read 1.75 m/s
Equate total energy before and after dive:
(1/2)mv²=(1/2)mx²+mgh
=(1/2)m*1.75² + mg(3)
Cancel out m and solve for v.
g=acceleration due to gravity = 9.81 m/s².