Bullet drop at 200m range with initial speed 250 m/s

[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!

But you're really interested in the time it takes for the horizontal distance traveled 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 traveled 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. So I opted to put it the way I did, without realising that it was wrong.