Question #92775

1 Answer
Jan 13, 2017

Answer:

#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.