A 72.0 kg swimmer jumps into the old swimming hole from a tree limb that is 4.00 m above the water.
Part A- Use energy conservation to find his speed just as he hits the water if he just holds his nose and drops in.
Part B- Use energy conservation to find his speed just as he hits the water if he bravely jumps straight up (but just beyond the board!) at 2.35 m/s .
Part C- Use energy conservation to find his speed just as he hits the water if he manages to jump downward at 2.35 m/s .
8.88
Part A: Well, if the swimmer just holds his nose and drops in, it means he's not really putting any effort, he's just letting gravity do its thing. So, let's crunch the numbers. Using energy conservation, we can say that the potential energy he loses is equal to the kinetic energy he gains.
So, his potential energy is mgh, where m is his mass (72.0 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height (4.00 m).
His kinetic energy is (1/2)mv^2, where v is the velocity we're trying to find.
Setting them equal, we get mgh = (1/2)mv^2.
Cancel out the mass, and solve for v: v = sqrt(2gh).
Plug in the values and get v = sqrt(2 * 9.8 m/s^2 * 4.00 m) = 8.86 m/s.
So, if he just holds his nose and drops in, he'll hit the water with a speed of 8.86 m/s.
Part B: Now, if he bravely jumps straight up (but just beyond the board!) at 2.35 m/s, he's got a bit of an initial velocity. We can still use energy conservation to solve this.
Again, we have mgh = (1/2)mv^2.
This time, he starts with some initial kinetic energy, since he jumps up. The energy conservation equation becomes mgh + (1/2)mv_initial^2 = (1/2)mv_final^2.
We can rearrange that equation to solve for v_final: v_final = sqrt(2gh + v_initial^2).
Plug in the values and get v_final = sqrt(2 * 9.8 m/s^2 * 4.00 m + (2.35 m/s)^2) = 9.15 m/s.
So, if he bravely jumps straight up at 2.35 m/s, he'll hit the water with a speed of 9.15 m/s.
Part C: Now, if he manages to jump downward at 2.35 m/s, he's got an initial velocity in the opposite direction. We can still use energy conservation to solve this one as well.
Using the same equation as before, mgh + (1/2)mv_initial^2 = (1/2)mv_final^2, we can solve for v_final.
In this case, the initial velocity is negative, so we need to subtract it in the equation. The equation becomes v_final = sqrt(2gh - v_initial^2).
Plug in the values and get v_final = sqrt(2 * 9.8 m/s^2 * 4.00 m - (2.35 m/s)^2) = 7.42 m/s.
So, if he manages to jump downward at 2.35 m/s, he'll hit the water with a speed of 7.42 m/s.
Remember, these calculations assume no air resistance or other forces acting on the swimmer during the jump.
To find the swimmer's speed just as he hits the water, we can use the principle of energy conservation. Energy is conserved when the swimmer jumps off the tree limb, so we can equate the initial potential energy to the final kinetic energy.
The initial potential energy (PEi) can be calculated using the formula PEi = m * g * h, where m is the mass of the swimmer, g is the acceleration due to gravity (9.8 m/s^2), and h is the height above the water.
Part A:
Since the swimmer just holds his nose and drops into the water, his initial speed (vi) is 0. Therefore, his initial kinetic energy (KEi) is 0.
When he reaches the water, his final potential energy (PEf) is 0, as he is at water level. This means that all his initial potential energy is converted into final kinetic energy (KEf).
Therefore, we can write the equation as:
PEi + KEi = PEf + KEf
m * g * h + 0 = 0 + ½ * m * vf^2
m * g * h = ½ * m * vf^2
Simplifying, we get:
g * h = ½ * vf^2
Now, we can solve for the final velocity (vf) by rearranging the equation:
vf^2 = 2 * g * h
vf = sqrt(2 * g * h)
Plugging in the given values (m = 72.0 kg, h = 4.00 m, g = 9.8 m/s^2) and calculating, we can find the result.
Part B:
In this case, the swimmer jumps straight up (but just beyond the board!). Let's assume the swimmer reaches a maximum height of H. At this maximum point, his velocity becomes 0. Therefore, we can again use the principle of energy conservation to find his speed just as he hits the water.
The initial potential energy (PEi) is still m * g * h, but this time we don't need to consider initial kinetic energy (KEi), as it is 0.
At the maximum height H, his potential energy (PEmax) is m * g * H, and his kinetic energy (KEmax) is again 0.
When he reaches the water, his final potential energy (PEf) is 0, and his final kinetic energy (KEf) is again ½ * m * vf^2.
So, the equation becomes:
PEi + KEi = PEmax + KEmax + PEf + KEf
m * g * h + 0 = m * g * H + 0 + 0 + ½ * m * vf^2
Simplifying, we get:
g * h = g * H + ½ * vf^2
Now, we can solve for the final velocity (vf) by rearranging the equation:
vf^2 = 2 * (g * h - g * H)
vf = sqrt(2 * (g * h - g * H))
Plugging in the given values (m = 72.0 kg, h = 4.00 m, H is the maximum height, g = 9.8 m/s^2) and calculating, we can find the result.
Part C:
In this case, the swimmer jumps downward with an initial velocity vi of 2.35 m/s. Again, we can use energy conservation to find his speed just as he hits the water.
The initial potential energy (PEi) is m * g * h, and the initial kinetic energy (KEi) is ½ * m * vi^2, as he has an initial downward velocity.
At the water's surface, his potential energy (PEf) is 0, and his final kinetic energy (KEf) is ½ * m * vf^2.
The equation becomes:
PEi + KEi = PEf + KEf
m * g * h + ½ * m * vi^2 = 0 + ½ * m * vf^2
Simplifying, we get:
g * h + ½ * vi^2 = ½ * vf^2
Now, we can solve for the final velocity (vf) by rearranging the equation:
vf^2 = 2 * (g * h + ½ * vi^2)
vf = sqrt(2 * (g * h + ½ * vi^2))
Plugging in the given values (m = 72.0 kg, h = 4.00 m, vi = 2.35 m/s, g = 9.8 m/s^2) and calculating, we can find the result.
Remember to double-check your calculations and use proper units when plugging in the values.
(1/2) m v^2 = m g h
(1/2) m v^2 = m g h + (1/2)m(2.3^2)
exactly the same. These trick questions should not be getting to you. If you need to think about it if he jumps up at 2.35 he will pass the end of the board at 2.35 on the way down.