Question

Question #92775

Answer 1

`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.