Easiest Physics Problem Ever Why Can't I Solve

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Homework Help Overview

The problem involves a diver performing a handstand dive from a height of 12 meters, with questions regarding the time to hit the water, velocities at different heights, and mean acceleration. The subject area pertains to kinematics and energy conservation principles in physics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate time and velocities using kinematic equations but expresses uncertainty about their results. Some participants question the necessity of mass in the calculations, while others clarify that mass cancels out in energy equations. There is also discussion about the relevance of time in the context of constant acceleration.

Discussion Status

The discussion is active, with participants providing insights and clarifications regarding the equations involved. There is a mix of interpretations regarding the need for time in the calculations, and some guidance has been offered about using energy conservation principles.

Contextual Notes

Participants note that the original poster was not given a mass, which led to confusion about its necessity in the calculations. The discussion also reflects on the assumptions made regarding the diver's initial conditions and the nature of the dive.

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Homework Statement


A diver performs a handstand dive from the 12m tower. Compute the following.
A. Time to hit the water
B. The divers velocity at 6m
C. The divers velocity at impact
D. What was the divers mean acceleration acting on the diver



Homework Equations


d=Vit + 1/2 a t^2
vf^2 = vi^2 + 2 a d
vf = vi + a t
a average = Change in velocity/Change in time


The Attempt at a Solution


A. i tried to solve for time, using equation 1 and got 1.56 seconds
B. then i tried to solve for velocity at 6m, used the same equation but used half the time and got like -3.87m/s but could I use the equation 3 and plug in half the time and the Vi as 0 and got -7.65 m/s
C. I just doubled both of those values?? to get the velocity at impact
D. I would take the final - initial divided by the time
 
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That's because you forgot a key equation:

mgh = 1/2 mv2

And life is simplified.
 
i wasnt given a mass...
 
You don't need the mass. Mass cancels from both sides of the equation. You could have derived it from your second equation though.

vf2 = vi2 + 2ad

You know that vi=0. Therefore,

vf2 = 2ad

1/2 vf2 = ad

d is just h, the height. a is just g, the acceleration due to gravity.

1/2 vf2 = gh

1/2 mvf2 = mgh

(This last equation is just a statement of the conservation of energy.)
 
youre a genius :)
 
but what is time? did I do that correctly?
 
Why do you need the time? The acceleration was constant throughout the entire process.
 
the time it took for the diver to hit the water. am i overlooking something. i feel like this was soooo easy, i feel dumb
 
Oh sorry. That was my mistake. I didn't see that your question called for that. Use the last equation.

vf = vi + at

You know that vi = 0 and a = g.
 
  • #10
thanks :)
 

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