Question

At an amusement park, a swimmer uses a water slide to enter the main pool. a. If the swimmer starts at rest, slides with negligible friction, and down a 3.25 m long slide that descends through a vertical height of 2.31m, what is her speed at the bottom of the slide? b. Find the swimmer’s speed at the bottom of the slide if she starts with an initial speed of 0.840 m/s.

Answer 1

Answer:

a)6.7m/S

b)6.8m/s

Explanation:

Hello ! To solve the point b you must follow the steps below

1.Draw the slide taking into account its length and height and find the angle from which the swimmer is launched (see attached image)

2. Find the horizontal velocity (X) and vertical (Y) components (see attached image)

3) for the third step we must remember that as in the slide there is no horizontal acceleration the speed in X will remain constant at the end of the swimmer's path (Vx = 0.59m / s)

4)

the fourth step is to remember that vertically there is constant acceleration called gravity (g = 9.81m / s ^ 2), so to find the speed at the end of the route we use the following equation

$Vfy= \sqrt{Vy^2+2gy}$

where    

Vfy= final verticaly speed    

Vy=initial verticaly speed=0.59m/S

g=gravity=9.81m/S^2

y=height of slide=2.31m

solving

$Vfy= \sqrt{Vy^2+2gy}\\Vfy= \sqrt{(0.59)^2+2(9.81)(2.31)}=6.77m/s$

The last step is to add the velocity components vectorally at the end of the route with the following equation

$V=\sqrt{Vfy^2+Vx^2} =\sqrt{6.77^2+0.59^2} =6.8m/s$

point A

taking into account the previous steps we can infer that as the swimmer starts from rest, the velocity (Vx=Vy=O) is zero, so we should only use the formula for constant acceleration movement.

$Vfy= \sqrt{Vy^2+2gy}$

vy=0

$Vfy=\sqrt{2gy}$

Vfy=$\sqrt{2(9.81)(2.31)}$=6.7m/s