Elementary yet fun fezziks

Projectile 1 when it returns to the same level from where both are thrown will have its velocity vector directed with an angle of 37° with the horizontal....... this configuration is the same as that of the initial configuration of projectile 2.... Hence from this point onwards the time taken to reach the ground is equal to t2 tthus the time difference is equal to the time of flight (ie. the time taken by 1 to reach the same level from where it was projected) hence t1-t2 = 2uSinθ/g = 60/9.8 = 6.122 NEAREST integer = 6

You can split it to look at just the vertical components. With both having the same angle and velocity the vertical components for both are the same. The vertical component is found with trigonometry by Sin(37)x50 which gives around 32. For it to reach it's peak the time is found by dividing start speed by acceleration (gravity) so 9.81 as it is decelerating to 0. This gives around 3 seconds. The upwards ball will take 3 seconds to peak and 3 seconds just to return to original position so will clearly be 6 seconds behind as takes six seconds to reach same "starting point"