Cahpter 3 physics?

Question

A rifle is aimed horizontally at a target 47 m away. The bullet hits the target 2.0 cm below the aim point.

What was the bullet's flight time?

What was the bullet's speed as it left the barrel?

Answer 1

I think the question is incomplete.

If you get other information try these solution

1.this question can be solved by using conservation of momentum law if u have rifle mass and speed.

2.this also can be solved calculating actual path traveled by bullet and taking usual speed of bullet.

Answer 2

let the rifle be at the origin.and let us take the x axis to positive towards the right and the positive y-axis upwards.

therefore,we have for projectile motion:

x=v cos theta *t,where v is the initial speed,t is the time taken by the bullet to hit the target and theta is the angle of projection.we know x=47 m.

we have:

47=v cos theta *t.

since the rifle is aimed horizontally,theta=0 and hence cos theta =1 and sin theta is =0

we get:

47=v*1*t=v*t...(1)

since the bullet hits the 2 cm=0.02m below the aim below,it has a final y displacement as y=-0.02 m taking into account our sign convention for the axes.

we get,by using y=v*sin theta*t +0.5*acceleration due to gravity(9.8m/s^2)*t^2;

-0.02=0-0.5*9.81*t^2(here we take the negative value as the bullet is going upwards against gravity)

solving for the positive value of t since time is never negative,

we get :t=0.063 seconds.

substitute into (1)

we get:

47=v*t

therefore,v=47/t=47/0.063=746.03m/s.

therefore,the answers are:

bullet's flight time:0.063 seconds.

bullets speed as it left the barrel:746.03m/s.