Initial speed of a thrown/falling stone

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Another question, I may have more. The reason I'm here is to do some homework due soon. I'm also here to learn, but the sooner the better.

A stone is dropped into a river from a bridge 43.9 m above the water. Another stone is thrown vertically down 1.00 s after the first is dropped. Both stones strike the water at the same time. (a) What is the initial speed of the second stone? (b) Plot velocity vs. time on a graph for each stone, talking zero time as the instant the first stone is released. I'm not sure if you could do a graph on here, but if you could explain in, that would be great!
Thanks!
 
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Anyone have ideas to get me started. I'm not sure what to do next.
 
Hi, you have 2 unknowns, the time and initial speed, so you need two equations to find them. Remember that the time for the second situation, t2 is given by (t1-1). You need equations of the form x=ut+1/2at^2.
 
What does the u stand for in your eqation? How can I get a (acceleration) solved?
 
u stands for the initial velocity, maybe you call it [itex]v_{i}[/itex]? a is going to be g=9.81ms^-2, as both stones are falling under gravity. This is the same for both, as after the second stone leaves the person's hand, it will accelerate at g, as the person's hand is no longer providing a force.
 
OK, but I still don't know the initial velocity. If I did your equation, would it look like this?
x=u(1.00)+1/2(9.8m/s^2)1.00^2
?
 
u is what you're trying to find. That would be part of your second equation, but it should be (t-1.00) not 1.00, where t is the time you get from the first equation (for the stone dropped from rest).
 
find the time taken for the first stone to hit the water.
x=u*t+1/2*a*t^2 where u=0ms^-1
then use that value to find time for the second stone. t2=(t1-1)
finally use this same equation again to find u.
x=u*t+1/2*a*t^2
 
hello i have a very similar question to this but what i don't understant is t(subscript)1-1 this equals 0... i don't get that?? soo time is zero?