(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A(n) 9 g bullet is ﬁred into a 396 g block

that is initially at rest at the edge of a frictionless table of height 0.9 m. The bullet remains

in the block, and after impact the block lands

2.3 m from the bottom of the table.

The acceleration of gravity is 9.8 m/s.

2. Relevant equations

ΔX = Vi(t) + (1/2)(a)(t^2)

m(Vf-Vi) = -m(Vf-Vi)

3. The attempt at a solution

I separated the equation into vertical and horizontal parts.

Horizontal: ΔX=2.3, a = 0 m/s/s

Vertical: ΔY= .9, a= -9.8 m/s/s, Vi = 0 m/s (because it is launched horizontally).

Using the kinematic equation ΔX = Vi(t) + (1/2)(a)(t^2) to solve for the vertical time:

t = 0.4285714286 seconds

Since the vertical time is equal to the horizontal time, I plugged this time into the same kinematic equation to solve for the horizontal Vi:

Vi = 5.366666667 m/s

Now, I know you need to use the collision equation to solve for the initial velocity in the bullet, but I do not know how to set it up...

I have this so far

9(5.366666667-vi)=-____?_____

Thank you for the help :)

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# Homework Help: Bullet in Block- 2D Projectiles w/ Motion

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