Help with kinematics in one dimension free fall

AI Thread Summary
A wrecking ball falls from rest after its cable breaks, taking 1.2 seconds to fall halfway to the ground. The discussion revolves around using kinematic equations to find the total fall time. The correct approach involves recognizing that the distance fallen from rest is half the product of acceleration and time squared. By applying the relationship between the time for half the fall and the total fall, the total time can be calculated as 1.2 seconds multiplied by the square root of 2, resulting in approximately 1.7 seconds. Understanding these kinematic principles is crucial for solving similar problems effectively.
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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