Agun with a muzzle velocity of 100 m/s is fired horizontally from a tower. neglecting air resistance, how far downrange will the bullet be 8 seconds later? g

313.6 m downward

Explanation:

The distance covered by the bullet along the vertical direction can be calculated by using the equation of motion of a projectile along the y-axis.

In fact, we have:

where

y(t) is the vertical position of the projectile at time t

h is the initial height of the projectile

is the initial vertical velocity of the projectile, which is zero since the bullet is fired horizontally

t is the time

a = g = -9.8 m/s^2 is the acceleration due to gravity

We can rewrite the equation as

where the term on the left, , represents the vertical displacement of the bullet. Substituting numbers and t = 8 s, we  find

So the bullet has travelled 313.6 m downward.

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west