A diver springs upward from a board that is three meters above the water. At the instant she contacts the water her speed is 9.40 m/s and her body makes an angle of 71.0° with respect to the horizontal surface of the water. Determine her initial velocity, both magnitude and direction.
2007-02-11 07:41:59 · 1 answers · asked by allyn_03 2 in Science & Mathematics ➔ Physics
First, compute the horizontal component of her velocity since this will be constant throughout the dive ignoring air resistance:
9.4*cos(71)
On the descent at the height of the board the diver will be at the same speed and angle as the take-off from the board.
The horizontal speed will be the same, but the kinetic energy will be diminished by m*g*3 from the water entry.
1/2m*v^2=1/2*m*9.4^2-m*g*3
the mass divides out
v^2=9.4^2-2*g*3
v=sqrt(9.4^2-2*g*3)
=5.43 m/s
The angle will be
Cos(th)=9.4*cos(71)/5.43
th=55.7
j
2007-02-13 12:22:06 · answer #1 · answered by odu83 7 · 0⤊ 0⤋