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High school physics help?
A boy runs off the edge of a vertical cliff above a pit with a horizontal initial speed of 5 m/sec. The cliff is 30 meters above the surface of the ocean. How far from the base of the cliff will the person splash into the ocean?
I am having trouble solving for time, which is only the first part of solving for the answer. Could you please show the steps to solve for the answer? Thank you
1 Answer
- 8 months agoFavorite Answer
To evaluate the time use:
sy=uyt-1/2gt^2 this reduces to s=1/2gt^2 because uy =0 (there is no initial vertical motion when the boy runs off the cliff it's all horizontal. sy=30
t = root[2sy/g]
t= root[(2 x 30)/9.81]
t = 2.47 s
sx = vx x t
sx = 5 x 2.47
sx = 12.36m
sx= 12m
The boy splashes into the ocean 12m from the base of the cliff.