A diver springs upward with an initial speed of 2.1 m/s from a 4.5 m board.
(a) Find the velocity with which he strikes the water. (Hint: When the diver reaches the water, his displacement is y = -4.5 m (measured from the board), assuming that the downward direction is chosen as the negative direction.)b. What is the highest point he reaches above the water?
(a) His kinetic energy will increase by M g H, where H = 4.5 m. Therefore
V2^2 = V1^2 + 2 g H
V1 = 2.1 m/s
(b) The increase in height at the highest point, h, is the elevation at which the potential energy increases by (1/2)MV1^2
2 g h = V1^2
Let's solve the problem step-by-step:
Step 1: Determine the downward direction as the negative direction. Since the displacement is measured as y = -4.5 m (below the board), we can assume downward direction is negative.
Step 2: Using the kinematic equation, we can find the final velocity (vf) of the diver when he strikes the water. The equation we can use is:
vf^2 = vi^2 + 2ay
where:
- vf is the final velocity of the diver (what we need to find)
- vi is the initial velocity of the diver (2.1 m/s)
- a is the acceleration due to gravity (-9.8 m/s^2) since it is pulling the diver down
- y is the displacement (position of the diver when he strikes the water = -4.5 m)
Step 3: Plug in the known values into the equation:
vf^2 = (2.1 m/s)^2 + 2(-9.8 m/s^2)(-4.5 m)
Step 4: Simplify the equation:
vf^2 = 4.41 m^2/s^2 + 88.2 m^2/s^2
vf^2 = 92.61 m^2/s^2
Step 5: Take the square root of both sides to find vf:
vf = √(92.61 m^2/s^2)
vf ≈ 9.63 m/s (rounded to two decimal places)
Answer to part (a): The diver strikes the water with a velocity of approximately 9.63 m/s.
Step 6: To find the highest point the diver reaches above the water, we can use the fact that at the highest point, the velocity is 0 m/s.
Step 7: We can use the same equation as before but this time, we solve for y (displacement).
vf^2 = vi^2 + 2ay
Step 8: Plug in the known values into the equation:
0 m/s = (2.1 m/s)^2 + 2(-9.8 m/s^2)y
Step 9: Solve for y:
0 m/s = 4.41 m^2/s^2 - 19.6 m/s^2 y
-4.41 m^2/s^2 = -19.6 m/s^2 y
Step 10: Divide both sides by -19.6 m/s^2 to isolate y:
y = -4.41 m^2/s^2 / -19.6 m/s^2
y ≈ 0.225 m (rounded to three decimal places)
Answer to part (b): The diver reaches a highest point of approximately 0.225 m above the water.
To find the velocity with which the diver strikes the water, we can use the equation of motion, specifically the equation for vertical displacement:
y = v₀y * t + (1/2) * a * t²
where:
y is the displacement (in this case, -4.5 m),
v₀y is the initial vertical velocity (in this case, 2.1 m/s),
t is the time taken to reach the water,
and a is the acceleration due to gravity (-9.8 m/s², considering downward direction as negative).
We need to solve this equation for t. Rearranging the equation, we get:
(1/2) * a * t² + v₀y * t + y = 0
This is a quadratic equation in terms of t. We can solve it using the quadratic formula,
t = (-b ± √(b² - 4ac)) / (2a)
where a = (1/2) * a, b = v₀y, and c = y.
Substituting the values, we have:
t = [ -2.1 ± √(2.1² - 4 * (1/2) * (-9.8) * (-4.5))] / (2 * (1/2) * (-9.8))
Simplifying further:
t = [ -2.1 ± √(4.41 - (-88.2))] / (-9.8)
t = [ -2.1 ± √(92.61)] / (-9.8)
Now we can calculate both roots of the equation:
t₁ = [ -2.1 + √(92.61)] / (-9.8) ≈ 1.29 s
t₂ = [ -2.1 - √(92.61)] / (-9.8) ≈ -0.35 s
Since time cannot be negative in this context, we discard t₂. Therefore, the time taken to reach the water is approximately 1.29 seconds.
Now, we can find the velocity at which the diver strikes the water using the equation:
v = v₀y + a * t
Substituting the values:
v = 2.1 + (-9.8) * 1.29
Calculating:
v ≈ -11.45 m/s
Therefore, the velocity with which the diver strikes the water is approximately -11.45 m/s (taking the downward direction as negative).
To find the highest point the diver reaches above the water, we can use the equation for vertical displacement once again. At the highest point, the diver's vertical velocity becomes zero. Thus, we have:
v = v₀y + a * t
Rearranging the equation for t:
t = (0 - v₀y) / a
Substituting the values:
t = (0 - 2.1) / (-9.8)
Calculating:
t ≈ 0.2143 s
Now, we can calculate the displacement at this time:
y = v₀y * t + (1/2) * a * t²
Substituting the values:
y = 2.1 * 0.2143 + (1/2) * (-9.8) * (0.2143)²
Calculating:
y ≈ 0.2437 m
Therefore, the highest point the diver reaches above the water is approximately 0.2437 m.