Respuesta :

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]

     

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