Asked by Stuck
"A diver springs upward with an initial speed of 1.8 m/s from a 3.0 m board. Find the velocity with which the diver strikes the water."
Help please!
Help please!
Answers
Answered by
bobpursley
Is initial energy (KE and PE) has to equal the KE at impact.
KEimpact=PEorig+KEorig
1/2 m vf^2=mgh + 1/2 m vi^2
solve for vf.
Notice mass divides out, you know h=3, vi=1.8m/s, and g is Earth's gravitation field (9.8N/kg)
KEimpact=PEorig+KEorig
1/2 m vf^2=mgh + 1/2 m vi^2
solve for vf.
Notice mass divides out, you know h=3, vi=1.8m/s, and g is Earth's gravitation field (9.8N/kg)
Answered by
Damon
How far above the board will he go?
v = Vi - 9.8 t
0 = 1.8 - 9.8 t
t = .184 seconds until he stops at the top
how far up above the board was that?
h = 0 + 1.8 t - 4.9 t^2
h = 1.8(.184) - 4.9(.184^2)
= .331-.165 = .166 meters above the board
so he falls 3+.166 = 3.166 meters
3.166 = 4.9 t^2
t = .804 seconds falling from top
v = 0 - 9.8 (.804)
= -7.88 meters/second
v = Vi - 9.8 t
0 = 1.8 - 9.8 t
t = .184 seconds until he stops at the top
how far up above the board was that?
h = 0 + 1.8 t - 4.9 t^2
h = 1.8(.184) - 4.9(.184^2)
= .331-.165 = .166 meters above the board
so he falls 3+.166 = 3.166 meters
3.166 = 4.9 t^2
t = .804 seconds falling from top
v = 0 - 9.8 (.804)
= -7.88 meters/second
Answered by
Stuck
Thank you! I got it!
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