Free Falling Objects: Find Initial Velocity of Q

[stupif]

1. stone p is releasedfrom a cliff of height 122.5m above the ground. after falling a distance of 19.6m, another stone Q is lauched straight downward.if given that p and Q reached the ground same time, find the initial velocity of Q( answer :26.1m/s

i using s=ut +1/2a[t][2] to find out the t, t= 5s
time of p=time of Q
i using s=ut +1/2a[t][2] to find out the initial velocity of u by change the S.
my answer is -3.92m/s

 

[uart]

Hi stupif. You're using the correct equations and your figure of 5 seconds is correct for the time of flight of the first stone.

The key to getting the correct answer is to figure out the correct time of flight for the second stone. So tell us what number you're using for this quantity and how you arrived at it.

 

[stupif]

then i used 122.5m - 19.6m=102.9m, so i using 102.9m as stoneQ's distance

 

[uart]

then i used 122.5m - 19.6m=102.9m, so i using 102.9m as stoneQ's distance

No you're interpreting the question incorrectly. The second stone is released from the same height as the first one (that is, 122.5m). This means that the second stone falls the same distance but does so in a shorter time period. If you work out what this shorter time period is then you can easily work out the required initial velocity. (the books answer is correct btw).

 

[stupif]

then s= same, t=same, a=same, just u is variable. how to calculate??

 

[uart]

No, "t" is NOT the same. The second stone is released AFTER the first stone has already fallen 19.6m, yet they arrive at the ground at the same time! So the time of flight of the second stone MUST be smaller. So that's the key to the problem, find the time of flight for the second stone and then you can easily find "u".

How long does it take for the first stone to fall 19.6m?

Use the above to find the time of flight for the second stone.

 

[stupif]

thank you...i got it...