Question #92775

Jan 13, 2017

$37.4 m {s}^{-} 1$, rounded to one decimal place.

Explanation:

It is supposed the bullet is fired from the ground level. When the bullet is fired it has velocities ${V}_{x} = 30 m {s}^{-} 1$ and ${V}_{y} = 40 m {s}^{-} 1$ in the $x \mathmr{and} y$ directions respectively. Ignoring the effect of air resistance there is no change in the velocity in $x$ direction.

The motion in the $y$ direction is governed by motion under the influence of acceleration due to gravity $= 9.81 m {s}^{-} 2$.
Kinematic equation is

$v = u + g t$
where $v$ is final velocity after time $t$ and $u$ is initial velocity.

Inserting given values we get, remember that direction of acceleration due to gravity is along $- y$ direction
${V}_{y 1.8} = 40 - 9.81 \times 1.8 = 22.342 m {s}^{-} 1$

Modulus of velocity of bullet at $t = 1.8 s$ is given by the expression
$| {V}_{1.8} | = \sqrt{{V}_{y 1.8}^{2} + {V}_{x}^{2}}$
$\implies | {V}_{1.8} | = \sqrt{{\left(22.342\right)}^{2} + {\left(30\right)}^{2}} = 37.4 m {s}^{-} 1$, rounded to one decimal place.