Kinematics Question (two objects thrown up)

[Kyleman]

I'm sure this is an easy question but I just can't seem to understand which 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.

 

[Vagrant]

Considering that you have initial velocities, accelerations and a relation between time taken, how about the second equation of motion?

 

[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

 

[Kyleman]

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?

 

[mbrmbrg]

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)

 

[Kyleman]

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.

 

[mbrmbrg]

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.

 

[KBark15]

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

 

[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

 

[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