How far does the object travel horizontally? Two-Dimensional Motion

[eioz]

Homework Statement

A rifle bullet is fired from the top of a cliff at an angle of 30° below the horizontal. The initial velocity of the bullet is 800 m/s. If the cliff is 80 m, high, how far does it travel horizontally?

θ = 30°
v₀ = 800 m/s
y = 80 m

Homework Equations

I may be totally off here with these equations (I'm part of an independent study program and their physics curriculum doesn't even have an online teacher, so I've been flying blind since day one), but I think these are relevant:

v_y = Vsinθ
|v| = √v_x² + v_y²

The Attempt at a Solution

v_y = 800sin30 = 400

800 - 400 = 400 = v_x

I then thought I could go from there to solve for x using x = x₀ + v_0xt + 1/2at², but I don't have anything for t, so I'm stuck.

 

[voko]

With 800 - 400 = 400 = vx you got the right answer, but only accidentally. To get the correct answer the correct way, you must use the second equation you listed in the relevant eq-s section.

Before you you compute the horizontal distance, ask yourself how far the bullet will travel vertically. And how long that will take.

 

[eioz]

I actually was using the equation I listed in the relevant eq-s section, but I do see how that isn't entirely clear.

I unfortunately really do not know where to go from where I'm at. The bullet will travel 80 m vertically, yes? But I don't understand how to use any of that to help me get time.

Could I use y = y₀ + v₀t to find out time, or does that not work here?

 

[voko]

The bullet will travel 80 m vertically, yes? But I don't understand how to use any of that to help me get time.

The distance of vertical travel depends on the initial vertical velocity, acceleration and time. You know all of these except the time, so you should be able to find it.

 

[Nickie]

To find the horizontal distance traveled, we can use the equation x = v0x * t, where v0x is the initial velocity in the x-direction and t is the time. In this case, the initial velocity in the x-direction is 800*cos30°, which is approximately 692.82 m/s. To find the time, we can use the equation y = y0 + v0yt - 1/2gt^2, where y0 is the initial height, v0y is the initial velocity in the y-direction (which we can calculate using v0y = v0*sin30°), and g is the acceleration due to gravity, which is approximately 9.8 m/s^2. Plugging in the values, we get 80 = 0 + 400t - 4.9t^2. Solving for t, we get t ≈ 0.318 seconds. Plugging this back into the first equation, we get x = 692.82 * 0.318 = 220.31 meters. Therefore, the bullet travels approximately 220.31 meters horizontally before hitting the ground.