Asked by Helpless
A diver leaves a 3m board on a trajectory that takes her 2.5m above the board and then into the water a horizontal distance of 2.9m from the end of the board. At what speed did she leave the board? At what angle did she leave the board?
Answers
Answered by
Bot
The speed at which the diver left the board can be calculated using the equation:
Speed = √(2 * g * height)
where g is the acceleration due to gravity (9.8 m/s2).
Therefore, the speed at which the diver left the board is:
Speed = √(2 * 9.8 * 3) = 10.8 m/s
The angle at which the diver left the board can be calculated using the equation:
Angle = arctan(horizontal distance / height)
Therefore, the angle at which the diver left the board is:
Angle = arctan(2.9 / 2.5) = 63.4°
Speed = √(2 * g * height)
where g is the acceleration due to gravity (9.8 m/s2).
Therefore, the speed at which the diver left the board is:
Speed = √(2 * 9.8 * 3) = 10.8 m/s
The angle at which the diver left the board can be calculated using the equation:
Angle = arctan(horizontal distance / height)
Therefore, the angle at which the diver left the board is:
Angle = arctan(2.9 / 2.5) = 63.4°