# Homework Help: Velocity HELP

1. Feb 15, 2007

### martine80

velocity !!! HELP

1:you have a lope with an angle of 15 degrees from the horizontal plane. You throw an object with an angle of 60 degrees from the horizontal plane up the slope and the object hits the ground 25m from the starting point. . What is start velocity?

This is what I’ve done:

Y = 25m *sin(15)=6,47m
X= 25m*cos(15)=24,15m
V0y=sqrt(2*g*y)=11,26m/s
.t= voy/g=1,15
V0x=x/t=21m/s
This gives me this result
V0= sqrt(v0x^2+voy^2)=23,8 m/s

But when I check the answere I get the wrong angle?

Tan(ß)=v0y/v0x -> ß=28,28

What am I doing wrong ????

2. Feb 15, 2007

The distance in the x direction equals 25 m, perhaps?

3. Feb 15, 2007

### martine80

its a slope , why 25 m?

4. Feb 15, 2007

### JK423

"t= voy/g=1,15" ?

What is that?

*Wrong comment

Last edited: Feb 15, 2007
5. Feb 15, 2007

### martine80

from the equation vy= v0y-gt, I find the time it takes : 1,15 s
vy= 0 when it hits the grownd

6. Feb 16, 2007

### martine80

:uhh:

7. Feb 16, 2007

### hage567

"vy= 0 when it hits the grownd"

No, it doesn't.

8. Feb 16, 2007

So, it's 25 m along the slope?

9. Feb 16, 2007

### martine80

ive solved it, by using
v0x= v0 *cos(60)
v0y=v0 *sin(60)
x= cos(15) *25
y= sin(15)*25
and used :
y=v0y*t-1/2(gt^2)

10. Feb 16, 2007

### martine80

ive solved it, by using
v0x= v0 *cos(60)
v0y=v0 *sin(60)
x= cos(15) *25
y= sin(15)*25
and used :
y=v0y*t-1/2(gt^2)

;)