Oh, My Brain!

[physicsss]

A shooter aims directly at a target on the same level 200 m away.

a) If the bullet leaves the gun at a speed of 250 m/s, by how much will it miss the target?

b) At what angle should the gun be aimed so the target will be hit?

[recon]

The bullet will travel horizontally at 250 m/s, thus making the nearest encounter with the target at 200/250 = 0.8 s. In this time, how much will the bullet have fallen from its original position? Remember that the acceleration due to gravity is 9.8 m/s2. The answer to (b) should immediately follow.

[physicsss]

For part a, I used the equation y2=y1+v1*t-1/2*g*t^2
since y1=0 and v1=0 in the y-direction, the equation is left with y=1/2*g*t^2
y=-1/2*(-9.8)*0.8^2=3.14 (is my sig figs correct?)

For b, I used v^2=v1^2-2gy, since at halfway distance is 100m and v=0 in the y direction, equation becomes

0=(sin(x)*250)^2+2*9.80*100

x=10.2
Is this and the sig figs right?

[Chronos]

Easier to apply the gravitational constant and triangulate using vectors. Try here and observe how the trajectory formula works.
http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra12

[physicsss]

nevermind. thanks for the link

[Tide]

The bullet will travel horizontally at 250 m/s, thus making the nearest encounter with the target at 200/250 = 0.8 s.

Actually, the nearest encounter will occur somewhat earlier than that! :smile:

But you're really interested in the time it takes for the horizontal distance travelled by the bullet to equal the original distance to the target.

[recon]

But you're really interested in the time it takes for the horizontal distance travelled by the bullet to equal the original distance to the target.


Yeah, that's what I meant. I just didn't know how to put it in words. :blushing: So I opted to put it the way I did, without realising that it was wrong.