At a local swimming pool, the diving board is elevated 21.5 ft above the pool's surface and overhangs the pool edge by L = 6 ft. A diver runs horizontally along the diving board with a speed of 13.2 ft/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 33% Part (a) Enter an expression for the time tw it takes the diver to fall off the end of the diving board to the pool's surface in terms of ,h, L, and g Grade Summary Deductions Potential 0% 100% Submissions Attempts remaining: 5 (10% per attempt) detailed view 12 3 0 Submit Hint Hints: 5% deduction per hint. Hints remaining: 4 Feedback: 0% deduction per feedback 33% Part (b) Calculate this time, tw, in seconds 33% Part (c) Determine the horizontal distance, aw in feet, from the edge of the pool to where the diver enters the water

Part (a) There are two motions involved: a horizontal motion and a vertical motion. The vertical motion is under gravity and is irrespective of the speed of the diver because the speed is horizontal. Also, the horizontal motion is not accelerated. Therefore, both motions take the same time [tex]t_w[/tex] which is determined by the vertical motion.

With respect to the vertical motion, there is no initial speed because the speed was fully horizontal. In fact, the diver could as well have dropped at the edge of the board and he would spend the same time to get to the pool surface.

Initial speed, u = 0 ft/s

Acceleration, a = g (it is purely under gravity)

Distance, s, = h

Time, [tex]t_w[/tex]

Applying one of the equations of motion which has the required parameters,

[tex]s=ut_w +\frac{1}{2}at_w^2[/tex]

[tex]h=0 +\frac{1}{2}gt_w^2[/tex]

[tex]t_w^2=\frac{2h}{g}[/tex]

[tex]t_w=\sqrt{\dfrac{2h}{g}}[/tex]

Part (b) Substituting the values, taking g = 32.2 ft/s[tex]{}^2[/tex]

[tex]t_w=\sqrt{\dfrac{2\times24. 5}{32.2}}=\sqrt{1.335}=1.16[/tex] s

Part (c) Total horizontal distance = horizontal distance due to diving + overhang (L = 6)

Horizontal distance due to diving is dependent solely on the horizontal component of the motion. This component has no acceleration, that is, has uniform speed of 13.2 ft/s.

This distance = speed [tex]\times[/tex] time

= 13.2 [tex]\times[/tex] 1.16 = 15.3 ft

Total horizontal distance = 15.3 + 6 = 21.3 ft