This question involves the concept of semi-projectile motion. It can be solved using the equations of motion in the horizontal and the vertical motion.
The bullet drops "2.76 m" by the time it reaches the target.
First, we will analyze the horizontal motion. We assume no air resistance, so the horizontal motion will be uniform. Hence, using the equation of uniform motion here to find the total time to reach the target:
[tex]s = vt\\\\t = \frac{s}{v}[/tex]
where,
s = distance = 75 m
v = velocity = 100 m/s
t = time = ?
Therefore,
[tex]t = \frac{75\ m}{100\ m/s}[/tex]
t = 0.75 s
Now, we will analyze the vertical motion of the bullet. We will use the second equation of motion in the vertical direction to find the height dropped by the bullet.
[tex]h = v_it+\frac{1}{2}gt^2[/tex]
where, Â
h = height dropped = ? Â
vi = initial vertical speed = 0 m/s
t = time interval = 0.75 s Â
g = acceleration due to gravity = 9.81 m/s²
therefore,
[tex]h = (0\ m/s)(0.75\ s)+\frac{1}{2}(9.81\ m/s^2)(0.75\ s)^2[/tex]
h = 2.76 m
Learn more about equations of motion here: Â
brainly.com/question/20594939?referrer=searchResults
The attached picture shows the equations of motion in the horizontal and vertical directions.
