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The ball projected upward:
Let time taken to reach the highest point = t
v = u - gt
0 = v1 - gt
t = v1/g
After that, the ball will fall down.
Let the height of cliff be h.
There are two possiblities here:
1) if t1 is less than t. Then by the time the first ball reaches its maximum height, the second ball is already thrown up.
Distance travelled by the second ball for time t - t1 is
s = ut -1/2gt^22 = v2(t - t1) - 1/2g (t - t1)^2
Remaining distance to be covered by both balls after time t' is h - s.
Let t2 be the time taken to cover remaining distance.
For first ball:
s1 = ut + 1/2 gt2^2 = 1/2gt2^2 ( u = 0)
For second ball:
Velocity after time t - t1 will be the initial velocity for this distance
v = u - gt = v2 - g(t - t1)
So, s2 = vt2 -1/2gt2^2
s1 + s2 = h - s
1/2gt2^2 + vt2 -1/2gt2^2 = h - (v2(t - t1) - 1/2g (t - t1)^2)
You can solve for t2 here and then the total time taken will be t + t2 = v1/g + t2
2) if t1 is greater than t'. Then, in that case, time taken for first ball to reach its maximum height will remain same. Now in this case,'
t2 = t + t'
t = t2 - t
Where t' is the time gap after the first ball reaches maximum height and second ball is thrown. So, for time t', first ball will start falling down.
Applying equations of motion,
v = u + gt = gt' = g (t2 - t)(after time t')
s = ut + 1/2gt'^2 = 1/2gt'^2 = 1/2g (t2 - t)^2
Again, the remaining distance to be covered by both the balls will be h - s.
Let t'' be the time taken to cover this distance.
Distance travelled by first ball , s 1 = vt'' + 1/2 gt''^2
Distance travelled by second ball, s2 = v2t'' - 1/2gt''^2
s1 + s2 = h -2
Substitute all the values and solve for t''. Total time taken = t2 + t''
Hope it helps!
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