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

`s_(y)=u_(y)t+(1)/(2)a_(y)t^(2)\\9.5=4.9t^(2)`

and,

`v_(x)=1.39a_(x)+2.5`

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+(1)/(2)a_(y)t^(2)`

So,

On putting the values we can say that,

`s_(y)=u_(y)t+(1)/(2)a_(y)t^(2)\\9.5=(0)t+(1)/(2)(9.8)t^(2)\\t^(2)=(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+(1)/(2)a_(y)t^(2)\\9.5=4.9t^(2)`

and,

`v_(x)=1.39a_(x)+2.5`