Posted by **slien** on Monday, June 19, 2006 at 7:47pm.

a projectile is launched at 35 degrees above the horizontal with an initial velocity of 120m/s. What is the projectile's speed 3 seconds later?

Figure the new vertical velocity at t=3. then the speed as equal to sqrt(Vv^2+ Vh^2)

The other way to do this is figure how high it travels in three seconds, then use the fact that

Kineticenergy=initialKE- change in PE.

- phys -
**Anonymous**, Thursday, September 26, 2013 at 3:09pm
cfghdf

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