Question

n amusement park water slide, people slide down an essentially frictionless tube. The top of the slide is 3.0m above the bottom where they exitthe slide, moving horizontally, 1.2mabove a swimming pool. What horizontal distance do they travel from the exit point before hitting the water

Answer 1

Horizontal distance will be 3.79 m

Step-by-step explanation:

We have given h = 3 m

Acceleration due to gravity
`g=9.8m/sec^2`

From conservation of energy we know that
`mgh=(1)/(2)mv^2`

`v=√(2gh)`

`v=√(2* 9.8* 3)=7.66m/sec`

Now initial vertical velocity = 0 m/sec

Initial horizontal velocity = 7.66 m/sec

Now height h = 1.2 m

From second equation of motion

`h=ut+(1)/(2)gt^2`

`1.2=0* t+(1)/(2)* 9.8* t^2`

t = 0.494 sec

Now horizontal distance = Horizontal speed × time = 7.66×0.494 = 3.79 m