Answer:
The distance covered along the horizontal axis is [tex]s = 360 \ m[/tex]
The distance covered along the vertical axis is [tex]h = 78.4 \ m[/tex]
Explanation:
From the question we are told that
The horizontal velocity is [tex]u_x = 90 \ m/s[/tex]
The time considered is [tex]t = 4 \ s[/tex]
Generally the distance which the bullet will cover in the horizontal direction for the given time is mathematically represented as
[tex]s = u_xt + \frac{1}{2}* at^2[/tex]
Here a is the acceleration along the horizontal direction and the value is [tex]a = 0 \ m/s^2[/tex]
So
[tex]s = 90 * 4 + \frac{1}{2}* * 0 * 4^2[/tex]
=> [tex]s = 360 \ m[/tex]
Generally the distance which the bullet will cover in the vertical direction for the given time is mathematically represented as
[tex]h= u_y t + \frac{1}{2}* gt^2[/tex]
Here [tex]u_y[/tex] is the initial velocity of the bullet along the vertical axis and the value is [tex]u_y = 0 \ m/s[/tex]
[tex]h = 0 * 4 + \frac{1}{2}* 9.8 *4^2[/tex]
=> [tex]h = 78.4 \ m[/tex]
