Question

At a local swimming pool, the diving board is elevated h = 5.5 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.a) express the time (tw) it takes the diver to move off the end of the diving board to the pool surface in terms of v0, h, L, and g.
b) calculate the time, tw, in seconds, it takes the diver to move off the end of the diving board to the pool surface.
c) determine the horizontal distance, dw, in meters, from the edge of the pool to where the diver enters the water.

Answer 1

Answer:

Part a)

$t = \sqrt{\frac{2h}{g}}$

Part b)

$t = 1.06 s$

Part c)

$L = 4.86 m$

Explanation:

Part a)

The height of the diving board is given as

$h = 5.5 m$

now the speed of the diver is given as

$v_0 = 2.7 m/s$

when the diver will jump into the water then his displacement in vertical direction is same as that of height of diving board

So we will have

$y = v_y t + \frac{1}{2}at^2$

$h = 0 + \frac{1}{2}gt^2$

$t = \sqrt{\frac{2h}{g}}$

Part b)

$t = \sqrt{\frac{2h}{g}}$

plug in the values in the above equation

$t = \sqrt{\frac{2(5.5 m)}{9.81}$

$t = 1.06 s$

Part c)

Horizontal distance moved by the diver is given as

$d = v_0 t$

$d = 2.7 \times 1.06$

$d = 2.86 m$

so the distance from the edge of the pool is given as

$L = 2.86 + 2$

$L = 4.86 m$