Ap physics b question (high school)

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A stone is dropped from a 75m building, and after falling 15m, a second stone is thrown downward with an initial velocity so that both stones hit the ground simultaneously. The first stone's final velocity is calculated to be 38.340 m/s, and it takes approximately 3.912 seconds to reach the ground. The second stone must reach the ground in about 2.161 seconds, leading to confusion about the calculations. The initial velocity of the second stone was incorrectly calculated, with a correct approach involving the equation x = x_0 + v_0 t + (1/2) a t^2. The expected initial velocity for the second stone is -24.34 m/s, indicating a misunderstanding in the application of kinematic equations.
zhanglshi
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a stone is dropped from a 75m building. when this stone has dropped 15m, a second stone is thrown downaward with an initial velocity such that the two stones hit the ground at the same time. what was the initial velocity of the second stone?

please use kinematic equations. that is all i know on this subject. thanks in advance.
 
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Why don't you give it a try? Use kinematic equations, of course.

Start by figuring out how long it takes for the first stone to reach the ground.
 
well the first stones velocity is 38.340m/s and the time is 3.912s. but idk where to go from there..
 
zhanglshi said:
well the first stones velocity is 38.340m/s and the time is 3.912s. but idk where to go from there..
OK. How long does it take for the first stone to reach 15 m? Then figure out how much time the second stone has to make it to the ground.
 
i am confused on something. i didntt understand the problem well enough so i restarted my work. for the 1st stone, is the final velocity 0? if so, then when i used the v=v(initial)-at, my equation was 0=0+(-9.8)(t) and i got t=0. that is obviously false but idk what i did wrong.. i am terribly confused ):
 
zhanglshi said:
i am confused on something. i didntt understand the problem well enough so i restarted my work. for the 1st stone, is the final velocity 0?
No, but its initial velocity is zero, since it is just dropped, not thrown.
 
but when the stone is at 0m, isn't its velocity 0?
 
zhanglshi said:
but when the stone is at 0m, isn't its velocity 0?
No. It just fell 75 m, so why would you think its velocity would be zero? (We're talking about just before it hits the ground.)
 
oh right. so i got t=2.161 for the2nd stone to hit the ground. i used thex=x(not)=v(not)^2+(1/2)at^2 and got v(not)=-28.654. but according to my answers sheet its supposed to be -24.34. i have no idea what i did wrong..
 
  • #10
zhanglshi said:
oh right. so i got t=2.161 for the2nd stone to hit the ground.
OK.
i used thex=x(not)=v(not)^2+(1/2)at^2
That should be:
x = x_0 + v_0 t + (1/2) a t^2
 
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