Answer:
Explanation:
Given
Distance of target d=75 m
velocity of bullet v=250 m/s
First taking horizontal motion
as there is no acceleration is horizontal direction therefore [tex]a_x=0[/tex]
[tex]d=ut+\frac{1}{2}a_xt^2[/tex]
where u=initial velocity
[tex]a_x[/tex]=acceleration
t=time
[tex]75=250\times t[/tex]
[tex]t=0.3 s[/tex]
In this time Vertical distance moved by Bullet is
[tex]y=u_yt+\frac{1}{2}a_yt^2[/tex]
here [tex]u_y=0[/tex]
[tex]a_y=g=9.8 m/s^2[/tex]
[tex]y=\frac{1}{2}\cdot 9.8\cdot (0.3)^2[/tex]
[tex]y=0.441 m[/tex]