At a local swimming pool, the diving board is elevated h = 9.5 m above the pool's surface and overhangs the pool edge by L = 2 m. A diver runs horizontally along the diving board with a speed of v0 = 2.7 m/s and then falls into the pool. Neglect air resistance. Use a coordinate system with the horizontal x-axis pointing in the direction of the diver’s initial motion, and the vertical y-axis pointing up. Let the y-axis be directed vertically down. Express the time t_w it takes the diver to move off the end of the diving board to the pool surface in terms of v_0, h, L, and g. Calculate the time, t_w in seconds, it takes the diver to move off the end of the diving board to the pool surface. Determine the horizontal distance, d_w in meters, from the edge of the pool to where the diver enters the water.

1) The time it takes the diver to move off the end of the diving board to the pool surface, , is approximately 1.392 seconds

2) The horizontal distance from the edge of the pool to where the diver enters the water, , is approximately 5.76 meters

Explanation:

1) The given parameters are;

The height of the diving board above the pool's surface, h = 9.5 m

The length by which the diving board over hangs the pool L = 2 m

The speed with which the diver runs horizontally along the diving board, v₀ = 2.7 m/s

Taking = The time it takes the diver to move off the end of the diving board to the pool surface

Therefore, we have from the equation of free fall;

h = 1/2 × g × ²

Where;

g = The acceleration due to gravity = 9.81 m/s²

Substituting the values, gives;

9.5 = 1/2 × 9.81 × ²

= √(9.5/(1/2 × 9.81)) ≈ 1.392 s

The time it takes the diver to move off the end of the diving board to the pool surface = ≈ 1.392 s

2) The horizontal distance, , in meters from the edge of the pool to where the diver enters the water is given as follows;

= L + v₀ × = 2 + 2.7× 1.392 ≈ 5.76 m

∴ The horizontal distance from the edge of the pool to where the diver enters the water ≈ 5.76 meters.