Diver Free Fall Physics

  • Thread starter RKNY
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  • #1
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Problem 1

Homework Statement


A diver springs upward with an initial speed of 1.8 m/s from a 2.5 m board.
(a) Find the velocity with which he strikes the water. (Hint: When the diver reaches the water, his displacement is y = -2.5 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?

Homework Equations


The four kinematics equations


The Attempt at a Solution


(a) used V^2 = V(initial)^2 + 2ay
V^2 = 1.8^2 + 2(-9.8)(-2.5) = 7.2 m/s which is wrong.
(b) Found out (b) to be 2.7 by 0 = 3.24 +19.6y and adding that to 2.5

Just can't seem to figure out (a). There is another problem which is similar and I guess I just don't seem to understand when something is shot/thrown up first and then falling.



Problem 2

Homework Statement


A ball is thrown upward from the top of a 55.0 m tall building. The ball's initial speed is 12.0 m/s. At the same instant, a person is running on the ground at a distance of 37.0 m from the building. What must be the average speed of the person if he is to catch the ball at the bottom of the building?

Homework Equations


The four kinematics equations

The Attempt at a Solution


The time the ball is thrown up
V = V(initial) + (a)(t)
0 = 12.0 + 9.8(t)
t = 1.2245 s
The time the ball is thrown down
y = V(initial)*t + 1/2(a)(t^2)
t = 3.35
Total time of 1.2245 s + 3.35 s = 4.6 s
Avg. Spd = D/T
37 m / 4.6 = 8.0 m/s = not the right answer
 
Last edited:

Answers and Replies

  • #2
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Problem 1
Do you know the work/energy theorem? If so that makes the first part very easy. If not, you know that on his return trip down, the instant he returns to the same height he jumped off that the velocity will be equal in magnitude (opposite in direction) to the initial velocity.

Problem 2
You made a sign error in your first equation, but recovered somehow. Anyway, I think you must misused the second equation in some way, but you never showed what numbers you put in.
 
  • #3
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Problem 1
Do you know the work/energy theorem? If so that makes the first part very easy. If not, you know that on his return trip down, the instant he returns to the same height he jumped off that the velocity will be equal in magnitude (opposite in direction) to the initial velocity.

Problem 2
You made a sign error in your first equation, but recovered somehow. Anyway, I think you must misused the second equation in some way, but you never showed what numbers you put in.
Problem 1
I tried using 1.8 as the V(initial), a = -9.80, y = -2.5
V^2 = -1.8^2 + 2(-9.80)(-2.5)
V = 7.23 but it was wrong. I don't know but it just seems right to me.

Problem 2
originally I put
y = -55, v(initial) = 0, a = -9.80
Y = 0 + 1/2(a)(t^2)
-55 = 0 + 1/2 (-9.80)(t^2)
t = 3.35029
 
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  • #4
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For the first one, did you consider what direction the velocity should be in?

For the second one, is the initial velocity really zero at the top of the building?
 
  • #5
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For the second one, is the initial velocity really zero at the top of the building?
Okay, thought about what u said and I believe I figured it out. Please lemme know if this is correct.

Time it takes for the ball to go up
V = V(initial) + (a)(t)
0 = 12.0 + (-9.80*t)
t = 1.2244 s
Time for the ball to go down from its peak being thrown up
0 = -12.0 + (9.80*t)
t = 1.2244 s
Time for the ball going down from the top of the building
V(initial) = -12.0
a = -9.80
y = 55

y = V(initial)*t + 1/2(a)(t^2)
55 = -12.0t - 4.9t^2
t = 4.7915 s

Add them all together
2(1.2244) + 4.7915 = 7.24 s

Avg. Speed = D/T

37/7.24 = 5.11 m/s avg. speed he needs to run to get to the building
 
  • #6
1,860
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Yes, that looks good. You could have saved yourself some time by multiplying the first time you found by two, since you know that it will take an object the same amount of time to return to its initial position from the max height as it took to get to the max height. But, if you are ever in doubt it is always best to do what you know works.
 

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