Help with kinematics in one dimension free fall

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Homework Help Overview

The problem involves a wrecking ball in free fall after its cable breaks, with a specific focus on calculating the time it takes to fall to the ground after falling halfway for 1.2 seconds. The subject area is kinematics in one dimension.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using different kinematic equations to find the time to fall from rest to the ground, with some questioning the validity of their calculations and others suggesting alternative methods to find the height and time.

Discussion Status

The discussion includes various attempts to solve the problem, with some participants providing equations and reasoning while others express confusion about their calculations. A potential approach involving ratios of distances and times has been suggested, but there is no explicit consensus on the final method to use.

Contextual Notes

Participants are working under the constraints of the problem statement and are exploring different interpretations of the kinematic equations. There is an indication of uncertainty regarding the application of these equations and the results obtained.

xlaserx7
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Homework Statement


A wrecking ball is hanging at rest from a crane when suddenly the cable breaks. The time it taktes for the ball to fall halfway to the ground is 1.2s. Find the time it takes for the ball to fall from rest all the way to the ground

Homework Equations


v=vo + at


The Attempt at a Solution


0 = 4.6 + (-9.80)t
t = - 5.2 s. time can't be negative so 5.2s
 
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Use the equation h = vo*t + 1/2*g*t^2
Using vo = 0 and t = 1.2s find the half height and then full height. From that find the time to fall from rest to the ground.
 
i don't understand why can't i use v=vo + at
and did i mess up on the time?
 
Last edited:
0 times 1.2 + .5 (-9.80)(1.44) = -7.056 gives me the height. full height is 7.056 times 2 = 14.112
now i am stuck again >.<

wait... now i use v=vo + at to find the final so

0 + (-9.80)1.2 = 11.76 = final velocity

then i use y = 1/2 ( v + v )t

y = 1/2(0 + -11.76)1.2
y = -7.056 omg i got the same thing?
 
Last edited:
Maybe look at it this way:

The distance dropped from rest is equal to 1/2 a*t2.

Exploiting this observe then that

\frac{H_1}{H_2} =\frac{1/2*g*t_1^2}{1/2*g*t_2^2}

This simplifies to - plugging in that H1 = H , and H2 = 2H

\frac{H}{2H} = \frac{1.2^2}{t_2^2} = \frac{1}{2}

t_2 = 1.2 * \sqrt{2}

Evaluating this equation looks remarkably easier to me.
 
ahh thank you
 
xlaserx7 said:
ahh thank you

So long as you learn the technique, you're welcome.

Good luck.
 

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