Question on the free-fall of 2 bodies

[Mr.maniac]

1. The probleas
statement, all variables and given/known data
Two stones are projected from the top of a tower 100m high each with velocity of 10m/s. One is projected upward while the other is projected downward. Find the ratio of velocity's by with which they strike the ground.

Homework Equations

S= ut +1/2(a)(t)^2
(v)^2-(u)^2=2as
v=u+at[/B]

The Attempt at a Solution

Let the stone projected upward be , body A
And the body projected downward be body B

For body A
(v)^2-(u)^2=2as
(v)^2-(100)=2(10)(100)
v=√(1900)
For body B before reaching max. height
(v)^2-(u)^2=2as
-(100)=(2)(10)(s)
5=s
For body B after reaching the max. height
(v)^2-(u)^2=2as
(v)^2=(2)(10)(105)
v=√(2100)

And I reached a dead end because
(10)(√21)/(10)(√19)

Ain't in the options
Please help

 

[PeroK]

1. The probleas
statement, all variables and given/known data
Two stones are projected from the top of a tower 100m high each with velocity of 10m/s. One is projected upward while the other is projected downward. Find the ratio of velocity's by with which they strike the ground.

Maybe you could do this without using any equations?

 

[Mr.maniac]

OK let me see

 

[Mr.maniac]

But the distance traveled by both the bodies is different

 

[PeroK]

But the distance traveled by both the bodies is different

Try describing the motion of the ball that gets thrown up. Do you notice anything?

 

[Mr.maniac]

OK wait a sec
Is it that the second body gains the same velocity as the first on whenn it is100m high

 

[PeroK]

OK wait a sec
Is it that the second body gains the same velocity as the first on whenn it is100m high

You mean if you throw a ball up at 10m/s, you know what speed it has when it comes back down to its starting point?

 

[Mr.maniac]

v=u+at (before reaching its highest point)
0=10+10t
1=t
Then it would take 1s to come back to its starting point and
v=u+at( after reaching the highest point)
v=10m/s
So basically both bodies start from the same point with the same vel. So the velocity before reaching the ground would be the same hence the answer is 1:1

 

[Mr.maniac]

Thanks for the help,

 

[PeroK]

v=u+at (before reaching its highest point)
0=10+10t
1=t
Then it would take 1s to come back to its starting point and
v=u+at( after reaching the highest point)
v=10m/s
So basically both bodies start from the same point with the same vel. So the velocity before reaching the ground would be the same hence the answer is 1:1

Yes, exactly. But, you may wish to go back to your equations and figure out how you got the wrong answer!

Hint: ##s## in the suvat equations is "displacement", not "total distance travelled".