Question

a bullet is fired horizontally at a velocity 10m/s from a watch tower 100m high. how long does it take to reach the ground?(g=9.8m/s^2)

Answer 1

Task. A bullet is fired horizontally at a velocity 10 m/s from a watch tower 100 m high. How long does it take to reach the ground? ($g=9.8m/s^{2}$)

Solution. First compute the time $t$ when the bullet will reach the ground. Let $v=(v_{x},v_{y})$ be the velocity of the bullet, where $v_{x}$ is the horizontal component of $v$ and $v_{y}$ is the vertical one. Since it is thrown horizontally, we see that

$v_{x}(0)=10m/s,\qquad v_{y}(0)=0$

Furhtermore, there is a gravitation force $F=mg$ acting on the bullet and it is directed along $y$-axis, whence the bullet moves with constant velocity along $x$-axis, i.e. $v_{x}$ does not depend on $t$:

$v_{x}(t)=10,$

and with constant acceleration $g$ along $y$-axis, so

$v_{y}(t)=-gt,$

and the height of the bullet is equal to

$h=100-\frac{gt^{2}}{2}.$

Hence the time $t$ when the bullet reach to the ground is satisfies the equation

$0=100-\frac{gt^{2}}{2},$

$t=\sqrt{200/g}=\sqrt{200/9.8}\approx 4.5175\,s.$

During this time the bullet will fly the distance

$d=v_{x}t=10\cdot 4.5175=45.175\,m.$

So the bullet will reach the ground at the distance 45.175 m from the tower.