Asked by claire
a ball rolls of from a cliff with a downward angle of 35 degrees. After 0.710seconds it lands 13m from the cliff. What is its velocity before it hits the ground?
Answers
Answered by
oobleck
The horizontal velocity is 13m/0.71s = 18.31 m/s
The vertical velocity as it left the cliff was v, where
v cos35° = 18.31
So, the vertical velocity was -v sin35° = -18.31 tan35° = -12.82 m/s
v(t) = -12.82 - 9.81*0.71 = -19.785 m/s
Now you have the vertical and horizontal components of the velocity upon impact.
The vertical velocity as it left the cliff was v, where
v cos35° = 18.31
So, the vertical velocity was -v sin35° = -18.31 tan35° = -12.82 m/s
v(t) = -12.82 - 9.81*0.71 = -19.785 m/s
Now you have the vertical and horizontal components of the velocity upon impact.
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