Question

At the local swimming pool, the diving board is elevated h = 8.5 m above the pool surface and overhangs the pool edge by L = 2 m. A diver runs horizontally along the diving board with a speed of v0 = 3.7 m/s and then falls into the pool. Neglect air resistance. Use a Cartesian coordinate system with its origin at the position of the diver just before falling. Let the direction that the diver falls be the negative y direction.
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)

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

Part b)

$t_w = 1.32 s$

Part c)

$d_w = 6.88 m$

Explanation:

Part a)

As we know that the diver will have zero velocity in vertical direction

so here we can say that

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

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

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

Part b)

as we know that

$h = 8.5 m$

$g = 9.81 m/s^2$

so we will have

$t_w = \sqrt{\frac{2(8.5)}{9.81}}$

$t_w = 1.32 s$

Part c)

Distance covered by diver from the edge of the pool is given as

$d_w = L + v_w t_w$

$d_w = 2 + (3.7)(1.32)$

$d_w = 6.88 m$