2. a 755N diver drops from a board 10.0 m above the water's surface. Find the divers speed 5.00 m above the water's surface. then find the diver's speed just before striking the water.
3. if the diver in item 2 leaves the board with an initial upward speed of 2.00 m/s find the divers speed when striking the water.
i don't know how to start these two at all...i don't need the answer i just need the formulas that i use for it...i don't understand which ones to use.. plz help
I don’t understand
I don’t understand can you show me
To solve problems involving motion, you can use the equations of motion, specifically the equation for free fall:
1) v = u + at
where:
- v is the final velocity
- u is the initial velocity
- a is the acceleration
- t is the time taken
Let's apply this equation to solve your problems step by step.
2) Finding the diver's speed 5.00 m above the water's surface:
First, we need to find the time it takes for the diver to fall from the board to a height of 5.00 m above the water.
To do this, we'll use the equation:
h = ut + (1/2)at^2
where:
- h is the height
- u is the initial velocity (which is 0 because the diver is initially at rest)
- a is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time taken
Rearrange the equation to solve for time (t) and substitute the given values:
5.00 = 0t + (1/2)(9.8)t^2
Simplify the equation to:
4.9t^2 = 5.00
Now solve for t by taking the square root of both sides:
t = sqrt(5.00 / 4.9)
Once you know t, you can use it to find the diver's speed 5.00 m above the water using the equation v = u + at. Since the diver is 5.00 m above the water, the height is 10.0 m - 5.00 m = 5.00 m. The acceleration due to gravity is -9.8 m/s^2 (negative because it acts downward). Substituting these values into the equation:
v = 0 + (-9.8)(t)
Calculate the value of v to find the diver's speed.
3) Finding the diver's speed just before striking the water:
To find the diver's speed just before striking the water, we can use a similar approach. However, this time we'll assume the initial velocity is 2.00 m/s upward. This means we need to find the time taken for the diver to fall from the board to the water's surface, which is a height of 10.0 m.
Using the same equation h = ut + (1/2)at^2 and substituting the given values:
10.0 = 2.00t + (1/2)(9.8)t^2
Simplify the equation and solve for t using the same steps as before. Once you know the time taken, you can find the diver's speed just before striking the water using the equation v = u + at, where the initial velocity u is 2.00 m/s and the acceleration a is -9.8 m/s^2.
By applying these equations and following the steps outlined above, you should be able to calculate the answers to both questions.
Use a conservation of energy approach for both problems. The CHANGE in (1/2)M V^2 equals the change in M g H.
That means the change in V^2 equals 2 g times the change in height above the water, H.
For the second problem, the water is 10 m below but he had a speed of V^2 = 2 m/s at the start.
Vfinal^2 = Vinitial^2 + 2 g h
= 4.0 + 2*9.8*10 = 200
V = 10 m/s