At the local swimming pool, the diving board is elevated h = 5.5 m above the poo

Question

At the local swimming pool, the diving board is elevated h = 5.5 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 = 4.6 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 g. (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. t_w = 0.43 (c) Determine the horizontal distance, d_w in meters, from the edge of the pool to where the diver enters the water.

Explanation / Answer

let's assume g = 10 m/sec^2
time tw to fall into water :
tw = 2h/g = 11/10 = 1.10 sec
Applying energy conservation :
mgh+1/2mVo^2 = 1/2mV^2
simplifying by m :
2gh+4.6^2 = V^2
V = 110+4.6^2 = 11.45 m/sec ( splashing speed)
Vx = 4.6 m/sec (horizontal splashing speed)
Vy = g*t = 10*1.1 = 11.00 m/sec (vertical splashing speed)
1 = arctan (Vx/Vy) = 22.69° diving angle related to the vertical
2 = 90-1 = 67.31 diving angle related to the horizontal
dw = Vo*tw = 4.6*1.1+2 = 7.06 m

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