Projectile motion rock throw speed

  • Thread starter Rasine
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  • #1
Rasine
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a rock is thrown from the tope pf a 20m buliding at an angle of 53 deg. above the horizontal. of the horizontal range of the throw is equal to the hieght of the buliding, with what speed was the rock thrown?

if i am given range i am trying to use x=xo+vot+.5at^2

but i don't know the time.

the velocity in the x direction would be vx=m/s and would equal v0cos53=20/s


in the y directio it would be vosin53=20/s

i don't know where to go...please help me
 

Answers and Replies

  • #2
denverdoc
961
0
well I think you are on the right path:

20=Vo(cos53) *t
0=20+V0(sin53)t+ .5(-9.8)t^2.
 
  • #3
hage567
Homework Helper
1,509
2
"if i am given range i am trying to use x=xo+vot+.5at^2"

This is not the equation you want for horizontal range. There will not be any acceleration in the horizontal direction once the rock is thrown. So how will that change things? The above equation will work for vertical distance, though.
So if you figure that out you will have two equations and two unknowns, allowing you to solve for v.

You will need to use the x and y components of the initial velocity. You are not quite right in your expressions for them, get rid of the 20/s (I'm not sure I understand what it is doing there like that) and just leave it equal to vy and vx to put into your other equations.

Try it and see if any of this helps.
 

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