Solve Physics Problem: Target 120 m Away, Speed 250m/s

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To solve the physics problem of a bullet aimed at a target 120 m away with a speed of 250 m/s, the relevant equations involve horizontal and vertical displacements. The time to reach the target can be calculated using t = Δx / v_o, yielding t = 0.48 seconds. During this time, the bullet falls vertically due to gravity, which can be calculated using Δy = 0.5 * a * t^2, resulting in a vertical drop of 1.12 m. Thus, the bullet will miss the target by 1.12 m as it travels horizontally while simultaneously falling. Proper attention to units and equations is essential for accurate calculations.
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The question says that a hunter aims directly at a target (on the same level) 120 m away. If the bullet leaves the gun at a speed of 250m/s, by how much will it miss the target?

So my values are v_o=250m/s, r=120m, and angle=0 since they're on the same level, right? And I need to know v to subtract from 120m.

I would just like to know what equation to use, not the answer to the problem. Thank you.
 
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The acceleration downward is 9.8 m/s2 so dv/dt= -9.8.
Integrating (or, since this is a constant, just mutiplying)
v= -9.8t+ initial vertical velocity= -9.8t+ 0 so v= dy/dt= -9.8t.
Integrating that, y= -4.9t2+ initial height= -4.9t2.

Neglecting air resistance (which we have to since there is no information on air resistance) there is no horizontal acceleration:
a= dv/dt= 0 so v= initial horizontal velocity= 250 m/s. and then
x= 250t. To go 120 m, requires that 250t= 120. Solve that equation for t and use that t in y= -4.9t2 to find how much the bullet has dropped.
 


To solve this physics problem, we can use the equation for horizontal displacement:

Δx = v_o * t

Where Δx is the horizontal displacement, v_o is the initial velocity, and t is the time.

In this case, we know that the initial velocity is 250m/s and the horizontal displacement is 120m. We can rearrange the equation to solve for t:

t = Δx / v_o

Plugging in the values, we get:

t = 120m / 250m/s = 0.48s

Now, we can use the equation for vertical displacement to find the vertical distance the bullet will travel during this time:

Δy = v_o * t + 0.5 * a * t^2

Where Δy is the vertical displacement, v_o is the initial velocity, t is the time, and a is the acceleration (which we can assume to be -9.8m/s^2 for simplicity).

Since we know that the initial vertical velocity is 0m/s (since the bullet is fired horizontally), we can simplify the equation to:

Δy = 0.5 * a * t^2

Plugging in the values, we get:

Δy = 0.5 * (-9.8m/s^2) * (0.48s)^2 = -1.12m

Therefore, the bullet will miss the target by 1.12m. This is because while the bullet is traveling horizontally for 0.48s, it is also falling vertically due to gravity.

I hope this helps and provides you with the necessary equation to solve the problem. Remember to always pay attention to the units and use the appropriate equations for the given situation.
 
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