Help with a probelm

  • Thread starter badman
  • Start date
57
0
A stone trown from the top of a 40m tower at an angle of 60 degrees above the horizontal hits the ground at the point 100m from the base of the tower.

find the speed at which the stone was thrown?

fine the speed of the stone just before it hits the ground? im sure i can get the answer to this with the help of the first.

it doesnt give you the time. :confused:
 
549
1
You'll need to calculate the time. For the first bit though, you don't need the time. Remember, you can consider horizontal and vertical motion separately. What can you say about the horizontal motion (assuming you're neglecting air resistance and stuff)?
 
well you know that the the time it takes for the stone to travel the 100m in the horizontal direction is the same amount of time that it travels in the vertical direction.
So ask yourself how long does it take the stone to travel 100m horizontally.
 
remember that there is no acceleration in the horizontal direction.
 
549
1
Grr, that's what I meant.
 
this is the displacement in the horizontal directon

[tex]x-x_0=v_0(\cos{\theta})t[/tex]

this is the displacemt in the vertical direction

[tex]y-y_0=v_0(\sin{\theta})t-\frac{1}{2}gt^2[/tex]

solve for time in the first equation and plug it into the second equation. :biggrin:
 
57
0
but what would V_0 equal too?
 
57
0
any one know ??
 
57
0
neone can give me some hints here.
 
2,208
1
You should develop the system of equations

[tex] 100 = v_0 \cos\left(\frac{\pi}{3}}\right) t [/tex]

[tex] -40 = v_0 \sin\left(\frac{\pi}{3}}\right) t - \frac{1}{2}gt^2[/tex]

Two equations, two unknowns.
 
57
0
hey thanks alot for ur time man, i already solved it. alot of work to twist the equations around.
 
show us your solution.
 
2,208
1
I got 29.x m/s and 6.x sec
 

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