## Kinematics Question (two objects thrown up)

I'm sure this is an easy question but I just can't seem to understand wich equations I need to use. A ball is thrown vertically upward with an initial speed of 11m/s. One second later, a stone is thrown veritcally upward with an initial speed of 25 m/s. Find the time it takes the stone to to catch up with the ball.

So right now I have for the ball: a= -9.80 m/s^2 initial V=11m/s initial y=0 t=t2-1 and for the stone a=-9.80m/s^2 initial V=25m/s initial y=0 t=?

So the way I was thinking is that the question is asking at what time the displacements are equal so I thought the equation I am supposed to use is (V^2-iV^2)/2a=(V^2-iV^2)/2a and solve for the V. This doesn't seem to work. Am I on the right track or just totally way off base? Any help or hints would be appreciated. Thanks.
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 Considering that you have initial velocities, accelerations and a relation between time taken, how about the second equation of motion?
 the movement 2 events equivalent displacement s1 = u1t1 + (1/2)gt12 s2 = u2t2 + (1/2)gt22 s1 = s2 u1t1 + (1/2)gt12 = u2t2 + (1/2)gt22 but substitute t1 by t2+1 (find t2) time of the second ball moving

## Kinematics Question (two objects thrown up)

 Quote by DAKONG the movement 2 events equivalent displacement s1 = u1t1 + (1/2)gt12 s2 = u2t2 + (1/2)gt22 s1 = s2 u1t1 + (1/2)gt12 = u2t2 + (1/2)gt22 but substitute t1 by t2+1 (find t2) time of the second ball moving
Could you tell me why it is t2+1 instead of t1+1? I'm just thinking that because the stone is thrown up one second later. As in ball=t stone=t+1 ? are you saying it's the other way around?

 Could you tell me why it is t2+1 instead of t1+1?
t2=stone's time
t1=ball's time

Because however long the stone has been flying (t2), the ball has been flying for one second longer (t1=t2+1).
If you prefer, you could also say that no matter how long the ball has been in the air (t1), the stone has been in the air one second less (t2=t1-1)

 Quote by mbrmbrg t2=stone's time t1=ball's time Because however long the stone has been flying (t2), the ball has been flying for one second longer (t1=t2+1). If you prefer, you could also say that no matter how long the ball has been in the air (t1), the stone has been in the air one second less (t2=t1-1)
I get a negative value for t if I do this and a positive value the other way.. is my algebra just off? Thanks for the help.
 I would assume that yes, your algebra is off. But if you show what you did, you'll probably get a more precise and accurate answer.
 so much for getting an answer on this stuff eh? just a bunch of posts to show how great they are at physics and that you arent
 I'm still having problems with this question where do i go from u1=11m/s u2=25m/s s1=s2 t1=(t2+1) u1t1 + (1/2)gt12 = u2t2 + (1/2)gt22 (11m/s)(t2+1)+(4.9m/s2)(t2+1)2=(25m/s)(t2+1)+(4.9m/s2)(t2+1)2

 Quote by bfolster16 I'm still having problems with this question where do i go from u1=11m/s u2=25m/s s1=s2 t1=(t2+1) u1t1 + (1/2)gt12 = u2t2 + (1/2)gt22 (11m/s)(t2+1)+(4.9m/s2)(t2+1)2=(25m/s)(t2+1)+(4.9m/s2)(t2+1)2
math is wrong you only insert the (t2+1) onto one side of the equation..
But (t1-1) works better also our g is - because we are throwing the ball against gravity

(11m/s)(t1)+(-4.9m/s2)(t1)2=(25m/s)(t1-1)+(-4.9m/s2)(t1-1)2
This is now corrected^^^
and you just FOIL out

11t+(-4.9t2)=25t-25+(-4.9t2)+9.8t+4.9
cancel (-4.9t2) because its on both sides and simplify for t