Calculating Block's Max Height from Impulse/Velocity

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SUMMARY

The maximum height of a 5kg block, which is impacted by a 10g bullet traveling at 1000m/s and exiting at 400m/s, is calculated to be 0.073m (7.3cm). The impulse imparted on the block is determined to be 6kgm/s, resulting in a final velocity of 1.2m/s for the block. The height is derived using the kinematic equation y = [(v2^2 - v1^2)/-2a], where acceleration due to gravity (a) is -9.8m/s².

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joex444
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I found this to be the hardest question on my last test:

A 5kg block is resting on two tables such that the middle isn't resting on a part of a table. A 10g bullet is fired at 1000m/s from directly below the block perpendicular. The bullet exits the block going 400m/s. What was the maximum height of the block?

Here's what I did, the impulse was the change in momentum of the bullet, which would be .01kg(1000m/s)-.01(400m/s) = 6kgm/s. That impulse must have been imparted on the block, which would give it a final velocity of 1.2m/s (6kgm/s / 5kg). From there, y = [(v2^2 - v1^2)/-2a] letting a = -g = -9.8 and v2 = 0, v1 = 1.2, and I got 0.073m = 7.3cm.

Did I go about this right?
 
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