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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.

Respuesta :

Answer:

Part a)

[tex]t = \sqrt{\frac{2h}{g}}[/tex]

Part b)

[tex]t = 1.06 s[/tex]

Part c)

[tex]L  = 4.86 m[/tex]

Explanation:

Part a)

The height of the diving board is given as

[tex]h = 5.5 m[/tex]

now the speed of the diver is given as

[tex]v_0 = 2.7 m/s[/tex]

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

[tex]y = v_y t + \frac{1}{2}at^2[/tex]

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

[tex]t = \sqrt{\frac{2h}{g}}[/tex]

Part b)

[tex]t = \sqrt{\frac{2h}{g}}[/tex]

plug in the values in the above equation

[tex]t = \sqrt{\frac{2(5.5 m)}{9.81}[/tex]

[tex]t = 1.06 s[/tex]

Part c)

Horizontal distance moved by the diver is given as

[tex]d = v_0 t[/tex]

[tex]d = 2.7 \times 1.06[/tex]

[tex]d = 2.86 m[/tex]

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

[tex]L = 2.86 + 2[/tex]

[tex]L  = 4.86 m[/tex]

A) 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 is; t_w = √(2h/g)

B) The time (tw) it takes the diver to move off the end of the diving board to the pool surface is; 1.06 s

C) The horizontal distance moved by the diver is; 4.862 m

Projectile Motion

We are given;

Height at which diving board is elevated; h = 5.5 m

Speed of the diver; v_o = 2.7 m/s

A) We will use Newton's second equation of motion which is that;

S = ut + ½at²

For the y-axis, we will use as;

h = (v_y × t) + ½gt²

Where v_y is velocity in the vertical direction which is zero. Thus;

h = 0 + ½gt²

h = ½gt²

gt² = 2h

t² = 2h/g

t_w = √(2h/g)

B) Since h = 5.5m, then;

t_w = √(2 × 5.5/9.8)

t_w = √1.122449

t_w = 1.06 s

C) The horizontal distance moved by the diver will be calculated from the formula;

d = v_o × t_w

d = 2.7 × 1.06

d = 2.862 m

Now, the horizontal distance, dw from the edge of the pool is;

dw = L + d

dw = 2 + 2.862

dw = 4.862 m

Read more about projectile motion at;https://brainly.com/question/11049671

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