Please please help me with this one. . . Even if you just explain it to me. That would be a great help.?

Question

I really need to understand this lesson. And I'm so sorry if I'm posting my assignment on Y!A. Sorry. Please please help me answer this and explain it.

Thank you so much!

-tepishane-

1.) A woman holds a hose of 0.8m above the ground such that the water shoots out horizontally. The water hits the ground at a point 2m away. What is the speed with which the water leaves the hose?

2.) A stone is projected vertically upward with a velocity of 15 m/s from the top of a hill 68 m high. Find the maximum height attained and the velocity with which it strikes the foot of the hill.

Answer 1

A)Time to hit the ground = time to travel horizontally

y-axis:

y=.5*g*t^2

.8m=.5*9.81m/s^2*t^2

solve for t

x-axis:

x=v*t

v=x/t

v=2/t

B)y-axis:

v=vi-g*t

At maximum height v=0 so:

0=vi-g*t

t=vi/g

t=15m/s/9.8m/s^2

t=1,53s

y=vi*t-5*g*t^2

y=15*1,53-.5*9.8*1.53^2

find y maximum

from y maximum, free fall for (y maximum+68m)

y=.5*g*t^2

solve for t (time to hit the ground)

v=g*t (v:velocity with which it strikes the foot of the hill)

Answer 2

Gravity acceleration is 9.8 m/s/s.

First, you calculate the time it takes for the water to fall .8m to the ground.

  d = 1/2 a t^2

.8 = 4.9 * t^2

t^2 = .8 / 4.9

t = .4 sec

then you know that in that time, the water travels 2m horizontally, so

2m/.4sec = 5m/s

your answer is 5 meters per sec