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2-D Motion

  • Thread starter martyk
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  • #1
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Greetings,
Here is my following question

1. During volcanic eruptions, chunks of solid rock can be blasted out of the volcano; these projectiles are called volcanic bombs. Figure 4-58 shows a cross section of Mt. Fuji, in Japan. (a) At what initial speed would a bomb have to be ejected, at angle θ0 = 35˚ to the horizontal, from the vent at A in order to fall at the foot of the volcano at B, at vertical distance h = 2.90 km and horizontal distance d = 9.00 km? Ignore, for the moment, the effects of air on the bomb's travel. (b) What would be the time of flight?



2. y = (tan theta)x - (gx^2) / 2(V0costheta0)^2



3. The above equation listed is the only thing that I can think of that works, usually constant acceleration equations aren't seem to getting anywhere for me. But anyways, I solved the equation to just the initial velocity (V0), just wanted you guys' opinions on whether this is the right approach to the problem

Thanks,
 

Answers and Replies

  • #2
jambaugh
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I would start with the two components of motion, x(t) and y(t). Recall the equations of motion are the position equals the initial position plus initial velocity times time plus half the acceleration times the square of the time. Since the only acceleration is gravity only the y component has that term.

Break the initial velocity into components using trigonometry. Since the angle given is off of the horizontal the horizontal component is proportional to the cosine and the vertical proportional to the sine. (In general just remember cosine is 1 when angle is zero so cosine is associated with the direction off of which the angle is measured.)

[tex] x(t) = x_0 + V_0 cos(35^o)\cdot t [/tex]
[tex] y(t) = y_0 + V_0 sin(35^o) \cdot t - (g/2)t^2[/tex]
where g is 9.8 m/s^2 is the gravitational acceleration.

You have two equations and two unknowns (V0 and t) since you know the initial and final positions. Solve and be sure to keep track of your units.
 

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