In the absence of air resistance, the horizontal and vertical motions of the bullet can be treated independently. The horizontal motion is unaffected by gravity, while the vertical motion is influenced by gravity.
Given: Horizontal velocity (v_x) = 300 m/s Horizontal distance (s_x) = 60 m
Since the horizontal motion is unaffected by gravity, we can use the formula:
s_x = v_x * t
where: s_x is the horizontal distance, v_x is the horizontal velocity, and t is the time.
Rearranging the formula to solve for time:
t = s_x / v_x
t = 60 m / 300 m/s
t = 0.2 s
Therefore, it will take 0.2 seconds for the bullet to travel the horizontal distance of 60 meters.
Now, let's calculate the vertical distance (s_y) dropped by the bullet during this time. Since the bullet is subject to the acceleration due to gravity (9.8 m/s^2), we can use the equation:
s_y = (1/2) * g * t^2
where: s_y is the vertical distance dropped, g is the acceleration due to gravity, and t is the time.
s_y = (1/2) * 9.8 m/s^2 * (0.2 s)^2
s_y = (1/2) * 9.8 m/s^2 * 0.04 s^2
s_y = 0.98 m
Therefore, the bullet will have dropped 0.98 meters vertically while traveling a horizontal distance of 60 meters in the absence of air resistance.