What Is the Speed of the Ball When Kicked Back in This Soccer Scenario?

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In the soccer scenario, Jack kicks the ball to John at a velocity of 5.40 m/s, and they are 6 meters apart. The time taken for the ball to reach John is calculated to be 1.11 seconds. After John kicks the ball back, the total time for the entire exchange is 2.6 seconds. The user struggles with the calculations to determine the speed of the ball when kicked back to Jack, indicating confusion in the process. The discussion highlights the need for clarity in solving kinematic equations related to the motion of the ball.
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Homework Statement


Jack and John are standing 6m apart from each other playing soccer. Jack kicks the ball to John and John kicks it back immediately. Jack kicked the ball with a velocity of 5.40 m/s and after 2.6 s he got the ball back. Suppose there's no deceleration of the ball when it's moving and that we can neglect the time for John to kick the ball. What is the speed of the ball when it's kicked back to Jack?


Homework Equations



V= D/T

The Attempt at a Solution


5.4 m/s = 6 /t = 1.11 seconds
1.11 + 2.6 = 3.71 seconds total
3.71 = ( (6/5.4) + (6/v) ) = 2.3

I'm doing something wrong, because I'm not getting the right answer.
thanks in advance
 
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OUstudent said:
5.4 m/s = 6 /t = 1.11 seconds
1.11 + 2.6 = 3.71 seconds total
1.11 is the time it takes for the ball to reach John. 2.6 is the total time.
 
I'm an idiot...

Thanks
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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