Solving Velocity Problem: Find the Start Velocity !

[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 ?

 

[radou]

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

 

[martine80]

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

its a slope , why 25 m?

 

[JK423]

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

What is that?

*Wrong comment

 

[martine80]

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

What is that?

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

 

[martine80]

 

[hage567]

"vy= 0 when it hits the grownd"

No, it doesn't.

 

[radou]

its a slope , why 25 m?

So, it's 25 m along the slope?

 

[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)

 

[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)

;)