Question

A bullet is fired straight up at 500 m/s
A) how far does it go before it stops and turns around
B) what is its velocity when it returns to starting point
C) how long does it take before it returns to starting point

Answer 1

a )\[v _{f}^2=v _{i}^2+2a \Delta y\] It is going up so acceleration is negative.

Answer 2

Whatever you get for the displacement of y, use it for b.

b) \[v _{f}^2=v _{i}^2+2a _{y}\Delta y\]
\[\Delta y \] would be negative since it is going down.
vi would zero since it is 0m/s at the top.

Answer 3

c.) Use \[v _{f}=v _{i}+a _{y} t\] both times and add it up.
To find first t, let vi=500m/s and vf=0m/s
To find second t, let vf= the value you got for b and vi=0m/s

Answer 4

or use \[x _{f}=x _{i}+v _{i}t+\frac{ 1 }{ 2 }at^2\]

Answer 5

I am getting 2 answers for how long does it take before it returns to starting point...51 sec or 102 secs

Answer 6

You add the time when it goes up and when it goes down.