How Far Should a Pilot Release a Bomb to Hit a Target?

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To determine how far a pilot should release a bomb to hit a target, the discussion focuses on the kinematic equations governing projectile motion. The airplane travels horizontally at 300 m/s from an altitude of 400 m, leading to a calculated horizontal distance of 2.7 km from the target. The key equations involve the vertical motion to find the time of fall (T) and the horizontal distance (x(T)). The vertical motion equation y(T) = -400 m and the horizontal motion equation x(T) = vx(0)T are used to solve for T and subsequently x(T). Understanding these equations is essential for accurately determining the release point for the bomb.
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An airplane flies horizontally with a speed of 300 m/s at an altitude of 400 m. Assume the ground is level. What horizontal distance from a target must the pilot release a bomb so as to hit the target? I know the answer is 2.7 km because it is or a review packet prior to a big test however, I do not know how my teacher got the problem.



Is it x=1/2at^2+vot?
I am confused!
 
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To pick appropriate kinematic equations, first take an inventory of what we know about the bomb:
[let's assert that (x,y,t) = (0,0,0) = position and moment in xy-plane when bomb is released]
vx(0) = 300 m/s
vx(t) = 300 m/s
vy(0) = 0 m/s
T = time from bomb release to target impact
y(T) = -400 m
ay = -9.8 m/s²
ax = 0 m/s²

What are we looking for:
T
x(T)

Is this clear so far?
 
Yes, this is clear. So we have do we do x=1/2at^2+vot?
 
Your x-axis equation boils down to x(T) = vx(0)T because ax is zero. Need to find T. Can you apply the y-axis version of your equation and see where it takes you?
 
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