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klio [65]
2 years ago
12

At a local swimming pool, the diving board is elevated h = 9.5 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 = 2.5 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.

Mathematics
1 answer:
beks73 [17]2 years ago
6 0
<h2>Answer:</h2>

s_{y}=u_{y}t+\frac{1}{2}a_{y}t^{2}\\9.5=4.9t^{2}

and,

v_{x}=1.39a_{x}+2.5

<h2>Step-by-step explanation:</h2>

In the question,

Taking the elevation of pool along the y-axis, and length of the board along the x-axis.

On drawing the illustration in the co-ordinate system we get,

lₓ = 2 m

uₓ = 2.5 m/s

and,

h_{y}=9.5\,m

So,

From the equations of the laws of motion we can state that,

s_{y}=u_{y}t+\frac{1}{2}a_{y}t^{2}

So,

On putting the values we can say that,

s_{y}=u_{y}t+\frac{1}{2}a_{y}t^{2}\\9.5=(0)t+\frac{1}{2}(9.8)t^{2}\\t^{2}=\frac{9.5}{4.9}\\t^{2}=1.93\\t=1.39\,s

Now,

The <u>equation of the motion in the horizontal</u> can be given by,

v_{x}=u_{x}+a_{x}t\\v_{x}=2.5+a_{x}(1.39)\\So,\\v_{x}=1.39a_{x}+2.5

<em><u>Therefore, the equations of the motions in the horizontal and verticals are,</u></em>

s_{y}=u_{y}t+\frac{1}{2}a_{y}t^{2}\\9.5=4.9t^{2}

and,

v_{x}=1.39a_{x}+2.5

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