A hunter aims directly at a target ( on the same level) 75.0 m away. (a) If the bullet leaves the gun at a speed of 180 m/s , by how much will it miss the target? (b) at what angle should the gun be aimed so as to hit the target?

Question

anonymous · Answer

consider the vertically downwards vector of the motion of the bullet
acceleration = 9.8 ms-2, displacement = s,
horizontal motion initial velocity = 180. distance = 75,  so time t = 75/180 secs
back to vertical motion
a = 9.8, s= ?, t = 75/180, u = 0
so s = ut + 0.5* 9.8 * t^2
s  = 0 + 4.9 * (75/180)^2
    = 0.85 m  = distance from the target

anonymous · Answer

$$s=ut+0.5at^2$$$$0=180sin \theta(75/180)+0.5(9.81)(75/180)^2$$$$\theta =0.65^0$$