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Physics freefall/velocity problem?
Two students are on a balcony 16.6 m above the street. One student throws a ball vertically downward at 13.7 m/s. At the same instant, the other student throws a ball vertically upward at the same speed. The second ball just misses the balcony on the way down. (Assume the positive direction is upward.) How far apart are the balls 0.800 s after they are thrown?
1 Answer
- AlgolLv 71 decade agoFavorite Answer
The ball thrown upward goes through a displacement in 0.800s of :
∆y₁ = v₀t + 0.5gt²
= (13.7m/s)(0.800s) + 0.5(-9.80m/s²)(0.800s)²
= 7.82m
The ball thrown downwards :
∆y₂ = (-13.7m/s)(0.800s) +0.5(-9.80m/s²)(0.800s)²
= -14.1m
Since we are interested in the distance the two balls are apart after 0.800s :
∆y₁ - ∆y₂ = 7.82m - (-14.1m)
= 21.9m
Hope this helps.
