# phys

posted by
**slien** on
.

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.