At the local swimming pool, the diving board is elevated h = 7.5 m above the poo

Question

At the local swimming pool, the diving board is elevated h = 7.5 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 v_0 = 2.2 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 y-axis be directed vertically down. (a) 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 t_w = (b) Calculate the time, t_w in seconds, it takes the diver to move off the end of the diving board to the pool surface. (c). Determine the horizontal distance, d_w in meters, from the edge of the pool to where the diver enters the water.

Explanation / Answer

The spped of diver is vo, and length of diving board is L, since speed is constant till before the fall Hence time taken to reach the end of board is t1 = distance/speed = L/vo

and time taken by diver to touch the surface of the pool after the fall is t2 can be calculated by considering motion in vertical direction, initial velocity u = 0, height dropped h, accleration a = g m/s2

using S = ut + 1/2at2

h = 0 + 1/2gt2

t2 = (2h/g)1/2

a) Hence tw = t1 + t2 = L/vo + (2h/g)1/2

b) using values of variables h = 7.5 m, L = 2 m, vo = 2.2 m/s, g = 9.81 m.s2

we get tw = 2/2.2 + (2*7.5/9.81)1/2 = 2.15 s

c) Horizontal distance travelled by diver dw = L + horizontal distance after the fall in t2 seconds

dw = 2 + speed*time

dw = 2 + 2.2*1.24 = 4.72 m

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