A bullet is fired horizontally with 20 m/s from top of a building 20 m high. When the bullet is 10 m above the ground incidentally it hits a bird. Find time taken to hit the bird?
2 Answers
Explanation:
For Physics or Mechanics you should learn the "suvat" equations for motion under constant acceleration:
#{: (v=u+at, " where ", s="displacement "(m)), (s=ut+1/2at^2, , u="initial speed "(ms^-1)), (s=1/2(u+v)t, , v="final speed "(ms^-1)), (v^2=u^2+2as, , a="acceleration "(ms^-2)), (s=vt-1/2at^2, , t="time "(s)) :} #
Horizontal Motion
The bullet will move in the horizontal direction under constant speed of
Vertical Motion
The bullet will move in the vertical direction under constant acceleration due to gravity. We will choose positive increasing in a downwards direction.
The initial vertical speed is
# { (s=,10,m),(u=,0,ms^-1),(v=,"Not Required",ms^-1),(a=,g,ms^-2),(t=,T,s) :} #
So we can calculate
# 10 = (0)(T)+1/2(g)(T^2) #
# :. 10 = 1/2gT^2 #
# :. gT^2 = 20 #
# :. T^2 = 20/g #
If we take
# T^2 = 20/9.8 => T = +- 1.428571 ... #
Hence
Explanation:
We will assume that the only force acting on the bullet after it is fired is that of gravity (ignore air resistance) and treat this as a simple projectile motion problem which can be solved with kinematics.
The bullet is fired with an initial horizontal velocity, but no initial vertical velocity. Therefore,
We also know that for a simple projectile, there is no horizontal acceleration, and under only the influence of gravity, the vertical acceleration is equal to
#-g=-9.8m/s^2# Note that
#g# , the gravitational acceleration constant, is a positive value. The negative sign indicates a downward acceleration, and is used in projectile motion problems to describe a falling object.
With values for
#y_f=y_i+cancel(v_(iy)Deltat)+1/2a_yDeltat^2#
#=sqrt((2(-10m))/(-9.8m/s^2)#
#=1.4 s#
