Yahoo Answers: Answers and Comments for Help in solving this physics question? [Physics]
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https://answers.yahoo.com/question/index?qid=20081216234300AAT96mn
From Anonymous
enUS
Tue, 16 Dec 2008 23:43:00 +0000
3
Yahoo Answers: Answers and Comments for Help in solving this physics question? [Physics]
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https://answers.yahoo.com/question/index?qid=20081216234300AAT96mn
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From phuong: just using kinetic equations for this problem....
https://answers.yahoo.com/question/index?qid=20081216234300AAT96mn
https://answers.yahoo.com/question/index?qid=20081216234300AAT96mn
Wed, 17 Dec 2008 08:25:43 +0000
just using kinetic equations for this problem.
y(final)=y(initial)+v(initial)t+1/2at^2
choose the origin on the ground
y(final)=0
y(initial)=50.0m
v(initial)=10.0m/s
a= g = 9.81m/s^2 since the acceleration due to gravity points down and the coordinate we choose points up.
then we have eqn:
0=50+10t4.9t^2
t=4.37 or t=2.33 (dont take this negative solution)
so after 4.37s, the stone will hit the ground or the bottom of the cliff
2, using another kinetic eqn for velocity
v=v(initial)+at=109.81(4.37)= 32.87m/s (the negative sign indicate the stone is moving downward)
3, i dont know if there is another way to calculate this part faster. Personally i'll calculate the highest point the stone will reach first
use the equation v(final)^2v(initial)^2 = 2*a*delta y
at the top, the velocity of the stone is 0
delta y = 10/(2*9.81) = 0.51m
so we have the total distance traveled by the stone = 0.51*2 + 50 = 51.02 m

From ferric: first I will address question 2. work done by ...
https://answers.yahoo.com/question/index?qid=20081216234300AAT96mn
https://answers.yahoo.com/question/index?qid=20081216234300AAT96mn
Wed, 17 Dec 2008 08:09:09 +0000
first I will address question 2. work done by gravity equals change in kinetic energy. (1/2)mv^2 + mgh=(1/2)mv^2 the LHS is the initial velocity and the RHS is the final velocity, notice that you do not know mass, but also notice that it cancels out, so
(1/2)m(10^2)+m(9.81)(50)=(1/2)mv^2
v=sqrt(100+(2)(50)(9.81))=32.88 m/s
to answer the other 2 questions we break the problem into 2 parts. The stone going up and then coming down. first going up...
we know that at its peak the velocity is 0, so this equation is usefull...
V(final)=V(initial)+A*T......... 0=10(9.81)T....T=1.02
now that we know the time to the peak height we use this equation...
D(final)=D(initial)+V*T+(1/2)A*T^2....
D=0+(10)(1.02)+(1/2)(9.81)(1.02^2)=5.1 m
total distance traveled is then= (2)(5.1)+50 = 60.2 m
now for the total time in the air....
D(final) = 50
50=0+(10)*T+(.5)(9.81)T^2
solving that quadratic equation gives you T = 4.4 s roughly