Calculating Velocity and Time for Vertically Thrown Stones

[Cummings]

A simple problem but i can't seem to get the answer i am told. I am thinking that i have mixed up my equations of motion

The question i have to answer is this.

A boy throws 2 stones verticly downward from a 50meter high cliff inot a calm pool of water.The 2 stones are thrown 1 second apart and the first stones initial velocity is 2m/s downwards. The 2 stones are seen to have hit the water at the same time.
a, How long after the release of the first stone do the 2 stones hit the water?
b, what was the initial velocity of the second stone?
c, what are the velocity's of the 2 stones at the instant they hit the water?

for a i used s=ut+.5a(t squared)
s = distance (50m) u = initial velocity of stone 1 (2m/s) a = gravity (9.8m/s2) and t is the time in seconds.
from that i found t = 2.99 = the right answer

for b, i tried using s=ut+.5a(t squared) again but for the second stone. So i uped the time by 1 second as it was released one second after the first so t = 3.99 s = 50m a = 9.8m/s2 and u is unknown
i got u = - 7.02 m/s

but i am told the speed is -15.4 m/s

a little help would be appreciated.

 

[quantumdude]

Originally posted by Cummings
for b, i tried using s=ut+.5a(t squared) again but for the second stone. So i uped the time by 1 second

There is your problem: You went in the wrong direction. The second stone travels 1 second less than the first. You have it traveling 1 second more.

You have to subtract 1 second, not add it.

edit: typo

 

[Cummings]

so if it was dropped 1 second later but hit at the same time, the time it spent in the air is actualy 1 second less?

i realize now that just because stone 1 took 2.99 seconds does not mean that stone 2 will take the same amount of time + 1 just because it was released a second after it. It seems clear that if it is dropped one second after it then it must take one second less time if it is to reach the ground at the same time.

All about visualising the problem. I knew it was the time that was stuffing me up but can't say i have had to visualise that before. Thanks.

 

[quantumdude]

Originally posted by Cummings
so if it was dropped 1 second later but hit at the same time, the time it spent in the air is actualy 1 second less?

Yes.