Question

If a bullet that leaves the muzzle of a gun at 255 m/s is to hit a target 107 m away at the level of the muzzle (2.0 m above level ground), the gun must be aimed at a point above the target. (Ignore any effects due to air resistance.) (a) How far above the target is that point? m (b) How far behind the target will the bullet strike the ground? m

Answer 1

given,

velocity at which muzzle leave the gun = 255 m/s

target is at 107 m away

a) time taken by the bullet to reach the target

time =
`(107)/(255)`

t = 0.42 s

during this time the bullet will fall by

`h = (1)/(2)gt^2`

`h = (1)/(2)* 9.8* 0.42^2`

h = 0.86 m

you should hit 0.86 m above the target.

b) the total height of the bullet above ground

2 + 0.86 = 2.86 m

time taken

`t = \sqrt{(2h)/(g)}`

`t = \sqrt{(2* 2.86)/(9.8)}`

t = 0.76 s

total distance the bullet will travel horizontally

s = v × t

s = 255 × 0.76

s = 193.8 m

so, it will land at 193.8 -107 = 86.8 m behind the target