Asked by joe
You're feeding the fish at the pond by the basefall field and decide to do some physics calculations. You notice that you release the fish food 1.80 m above the surface of the water, and during one throw, when the food left your hand at 2.80 m/s with a direction of 11.0o below the horizon, you managed to hit a fish near the surface. How fast was the food going right before it hits the fish?
Answers
Answered by
oobleck
Vy = 2.80 sin11° - 9.8t
Vx = 2.80 cos11°
V = √(Vx^2 + Vy^2)
so how long did it take to hit the water?
1.80 + 2.80 sin11° t - 4.9t^2 = 0
use that t to find Vy
Vx = 2.80 cos11°
V = √(Vx^2 + Vy^2)
so how long did it take to hit the water?
1.80 + 2.80 sin11° t - 4.9t^2 = 0
use that t to find Vy
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