Calculating velocities with only acceleration and time

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To calculate the initial and final velocity of a ball thrown upward and caught at the same level, the relevant equations are Vf = Vi + at and delta x = Vi*t + (0.5)(t^2). Given the acceleration due to gravity is -9.8 m/s² and the time is 2.12 seconds, the initial and final velocities can be determined. Since the ball is caught at the same horizontal level, the initial and final velocities are equal in magnitude but opposite in direction. The discussion emphasizes the need to identify the correct equations and understand the relationship between velocity, acceleration, and time.
noelani
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This was a basic physics experiment. A ball was thrown upward and then caught. Calculate initial and final velocity given: acceleration= -9.8m/s^2 (gravity); time= 2.12sec



delta x=Vi*t+(.5)(t)^2
Vf=Vi+at




I actually don't know where to begin. I don't know an equation for velocity that doesn't require displacement (which I also need to find).
 
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if it's caught at the same horizontal level, then vi=vf [note the direction of each, one is + and the other is -]

solve!


ps. welcome to PF :)
 
drizzle said:
if it's caught at the same horizontal level, then vi=vf [note the direction of each, one is + and the other is -]

solve!


ps. welcome to PF :)

Thanks for the help. So how would I figure out which equation to use?
 
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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