The bullet travels a horizontal distance of 276.5 m
The bullet is shot forward with a horizontal velocity [tex]u_x[/tex]. It takes a time t to fall a vertical distance y and at the same time travels a horizontal distance x.
The bullet's horizontal velocity remains constant since no force acts on the bullet in the horizontal direction.
The initial velocity of the bullet has no component in the vertical direction. As it falls through the vertical distance, it is accelerated due to the force of gravity.
Calculate the time taken for the bullet to fall through a vertical distance y using the equation,
[tex]y=u_yt+\frac{1}{2} gt^2[/tex]
Substitute 0 m/s for [tex]u_y[/tex], 9.81 m/s²for g and 1.5 m for y.
[tex]y=u_yt+\frac{1}{2} gt^2\\ 1.5 m=(0m/s)t+\frac{1}{2} (9.81m/s^2)t^2\\ t=\sqrt{\frac{2(1.5m)}{9.81m/s^2} } =0.5530s[/tex]
The horizontal distance traveled by the bullet is given by,
[tex]x=u_xt[/tex]
Substitute 500 m/s for [tex]u_x[/tex] and 0.5530s for t.
[tex]x=u_xt\\ =(500m/s)(0.5530s)\\ =276.5m[/tex]
The bullet travels a distance of 276.5 m.
