Question #250f9

1 Answer
Apr 1, 2016

Answer:

Four times.

Explanation:

Let's assume that second body is thrown with a velocity #vecu# making an angle #theta# with #x# axis.

Velocity along #x#-axis #= u_x = u cos theta#
Acceleration along #x#-axis #a_x = 0#
Velocity along #y#-axis #= u_y = u sintheta#
Acceleration along #y#-axis #a_y = -g#

To obtain total Time of flight #t#, only #y# component of velocity needs to be considered. The ball rises up, comes to a stop and then starts its decent downwards. Once it reaches the ground level, magnitude of the #y# component of velocity is same when it was thrown. Only the direction is opposite. Using the equation

#v=u+at#, and inserting given values
#-usin theta=u sintheta-g t#

#t=2usintheta/g#.........(1)
Horizontal Range# = "Horizontal velocity" xx "Time of flight"#
#R= u cos theta xx 2 u sin theta/g#
#= u^2/g 2 sin thetacos theta#
#=(u^2 sin 2theta)/g# ....(2)

It is given that velocity of first body is double that of the second body and angle #theta# is same
From (2) we see that for the same angle of throw Range #Rprop u^2#.
#=># Distance covered by first body is #2^2=4# times the range of second.