A diver springs upward with an initial speed of 1.6 m/s from a 3.5-m board.
(a) Find the velocity with which he strikes the water. [Hint: When the diver reaches the water, his displacement is y = -3.5 m (measured from the board), assuming that the downward direction is chosen as the negative direction.]
To find the velocity with which the diver strikes the water, we can use the equation of motion:
velocity² = initial velocity² + 2 * acceleration * displacement
In this case, the initial velocity of the diver is 1.6 m/s, the acceleration due to gravity is -9.8 m/s² (taking downward direction as negative), and the displacement is -3.5 m (since the diver is moving downward).
Plugging these values into the equation, we get:
velocity² = (1.6 m/s)² + 2 * (-9.8 m/s²) * (-3.5 m)
Simplifying this equation:
velocity² = 2.56 m²/s² + 68.6 m²/s²
velocity² = 71.16 m²/s²
To find the velocity, we take the square root of both sides of the equation:
velocity = √(71.16 m²/s²)
Using a calculator, we find:
velocity ≈ 8.43 m/s
Therefore, the velocity with which the diver strikes the water is approximately 8.43 m/s.
vf^2=Vi^2+2*g*h
solve for Vf