Question #92775

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Question #92775

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Answer 1 · 2017-01-13T19:58:50

$37.4ms^-1$ , 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 $V_x = 30ms^-1$ and $V_y = 40ms^-1$ in the $x and y$ 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 $=9.81ms^-2$ .
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
$V_(y1.8)=40-9.81xx 1.8=22.342ms^-1$

Modulus of velocity of bullet at $t=1.8s$ is given by the expression
$|V_(1.8)|=sqrt(V_(y1.8)^2+V_x^2)$
$=>|V_(1.8)|=sqrt((22.342)^2+(30)^2)=37.4ms^-1$ , rounded to one decimal place.

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Question #92775

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