Respuesta :
Answer
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 = [tex]\dfrac{107}{255}[/tex]
t = 0.42 s
during this time the bullet will fall by
[tex]h = \dfrac{1}{2}gt^2[/tex]
[tex]h = \dfrac{1}{2}\times 9.8\times 0.42^2[/tex]
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
[tex]t = \sqrt{\dfrac{2h}{g}}[/tex]
[tex]t = \sqrt{\dfrac{2\times 2.86}{9.8}}[/tex]
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
