A diver springs upward from a board that is 3.90 m above the water. At the instant she contacts the water her speed is 14.9 m/s and her body makes an angle of 60.3 ° with respect to the horizontal surface of the water. Determine her initial velocity, both (a) magnitude and (b) direction.
First, consider energy.
Final KE=Initial PE+initial KE
1/2 m *14.8^2=mg*3.9+1/2 m v^2
solve for vi. That gives you magnitude.
now, consider horizontal velocity. It remains constant.
intial horizontal velocity=final horizVelocity
Vih=14.8*cos60.3 solve that.
angleinitial=arcCos(Vih/Vi) solve.
To determine the diver's initial velocity, we can break it down into horizontal and vertical components.
(a) Magnitude of the initial velocity:
The horizontal component of the velocity remains constant throughout the diver's motion since there is no horizontal acceleration. Therefore, the magnitude of the horizontal component of the initial velocity is the same as the magnitude of the horizontal component of the velocity just before hitting the water.
Given that the speed of the diver just before hitting the water is 14.9 m/s and the angle her body makes with respect to the horizontal surface of the water is 60.3°, we can use trigonometry to find the magnitude of the horizontal component of the velocity.
Horizontal component = velocity × cos(angle)
Horizontal component = 14.9 m/s × cos(60.3°)
Horizontal component ≈ 14.9 m/s × 0.5
Horizontal component ≈ 7.45 m/s
Thus, the magnitude of the initial velocity is approximately 7.45 m/s.
(b) Direction of the initial velocity:
The direction of the initial velocity will be the same as the direction of the horizontal component of the velocity. In this case, since the diver springs upward from a board, the horizontal component of the velocity will be directed horizontally.
Therefore, the direction of the initial velocity is horizontally.
To summarize:
(a) The magnitude of the initial velocity is approximately 7.45 m/s.
(b) The direction of the initial velocity is horizontally.