Rock's Max Horizontal Distance from Volcano: 500m

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SUMMARY

The maximum horizontal distance a rock can travel when ejected from a 500m high volcano at a speed of 30 m/s is calculated using projectile motion equations. The relevant equations include the range formula, range = [(V initial)^2 x (sin 2A)]/[g], and the vertical motion formula, y = yo + v*t - 0.5*g*t^2. The angle of projection should be 45 degrees for optimal range, and the gravitational acceleration is 9.8 m/s². The initial calculations must account for the height of the volcano to determine the correct horizontal distance.

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peace1709
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A 500m high volcano eject rocks at a speed 30 m/s.What is max horizontal distance reached by the rock?
 
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Hi peace1709, Welcome to PF.
What do you about the problem?
In what direction the rocks are moving?
which are the relevant equations?
 
i use the equation range= [(V initial)^2 x (sin 2A)]/[g].
where v inital is 30 m/s and i assume A=45 degrees, and g=9.8 m/s^2
but i got the answer wrong.
 
i got 91.7 m and its wrong, can u help me?
 
In your calculation you have not taken into account the height of the volcano.
 
oh ya.but i don't know how to relate the height. can u help me?
 
You can use the formula
y = yo + v*t - 0.5*g*t^2.
Here v = vo*sinθ
And t = x/vo*cosθ
y is the final position of the rock and yo is the initial position of the rock.
 

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