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.

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.