20 m/s that makes an angle of 35° with the horizontal. What is the vertical velocity of the projectile after 3 seconds? Take the upward direction to be positive.
v0y=v⋅sin35°=20⋅sin35°=11.47(m/s)v_{0y}=v\cdot \sin35°=20\cdot \sin35°=11.47(m/s)v0y=v⋅sin35°=20⋅sin35°=11.47(m/s)
vy(t)=v0y−gt→vy(3)=11.47−9.81⋅3=−17.96(m/s)v_y(t)=v_{0y}-gt\to v_y(3)=11.47-9.81\cdot 3=-17.96(m/s)vy(t)=v0y−gt→vy(3)=11.47−9.81⋅3=−17.96(m/s) . Answer
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