Question

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.

Answer 1

Answer:

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

and,

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

Step-by-step explanation:

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 equation of the motion in the horizontal can be given by,

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

Therefore, the equations of the motions in the horizontal and verticals are,

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

and,

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