Question

A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling horizontally. As the drawing shows, one person hits the water 5.00 m from the end of the slide in a time of 0.672 as after leaving the slide. Ignore friction and air resistance, find the height H in the drawing.

Answer 1

Answer:

2.79m

Explanation:

A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling horizontally. As the drawing shows, one person hits the water 5.00 m from the end of the slide in a time of 0.672 as after leaving the slide. Ignore friction and air resistance, find the height H in the drawing

height of the slide +distance fallen to the water will give us the height H

horizontal speed

distance=speed /time

5=speed/0.672

speed=.672*5

speed=3.36m/s

from conservation of energy , we say tat potential energy equals kinetic energy

(1/2)mV^2 = m g h

0.5*3.36^2/9.81

h1=.575m  eit of te slide

now to get the distance of the fallen water

s=ut +1/2at^2

u=0

s=0.5*9.818*0.672^2

s=2.21m

the total height is 2.21+0.575

2.79m

Answer 2

Answer:

H = 5 m

Explanation:

As the person leaves the slide horizontally so the time taken by the person to hit the water is given as

$t = 0.672 s$

so we can find the vertical velocity by which person will hit the water using kinematics

$v_y = u_y + at$

$v_y = 0 + (9.81)(0.672)$

$v_y = 6.6 m/s$

now the speed of the person at the end of the slide is given as

$v = \frac{L}{t}$

$v = \frac{5}{0.672}$

$v_x = 7.44 m/s$

now by energy conservation we can find the initial height

$mgH = \frac{1}{2}m(v_x^2 + v_y^2)$

$H = \frac{1}{2g}(v_x^2 + v_y^2)$

$H = \frac{1}{2(9.81)}(7.44^2 + 6.6^2)$

$H = 5 m$