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.
