a) The bullet will not reach the target
The motion of the bullet follows a parabolic path. First, we have to determine the time it takes for the bullet to reach the ground. We can do it by considering the vertical motion only, which is a free-fall motion, so we can use the equation
[tex]s=ut+\frac{1}{2}at^2[/tex]
where, taking downward as positive direction,
s = 1.5 m is the vertical displacement of the bullet
u = 0 is its initial vertical velocity
t is the time
[tex]a=g=9.8 m/s^2[/tex] is the acceleration of gravity
Solving for t,
[tex]t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(1.5)}{9.8}}=0.553 s[/tex]
So, the bullet lands 0.553 s after being shot.
The bullet is fired horizontally at a speed of
[tex]v_x = 1000 m/s[/tex]
So, the horizontal distance covered during this time is
[tex]d=v_x t = (1000)(0.553)=553 m[/tex]
And since the target is 700 m away, the bullet will not reach it.
b) 553 m
As we stated in the previous part, the bullet takes
t = 0.553 s
To land to the ground.
Also, it travels at
[tex]v_x = 553 m/s[/tex]
Therefore, it lands on the ground at a distance of
[tex]d=v_x t=(1000)(0.553)=553 m[/tex]
