Dr Gulshan S. answered 09/06/17
Tutor
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Physics Teaching is my EXPERTISE with assured improvement
Hi Zainab
Time taken by first packet to hit the ground is t
H = 1/2 gt 2 as initial velocity is zero
Using g = 10 m/s 2
800 = 5t 2
Gives t= 12.65 sec
Now to know where is second packet after
time available to second packet is 12.65- 1.0 sec = 11.65 sec
So the distance covered in this time
H' =1/2 gt 2 = 678.6 m
So the separation between the two = 800.0-678.6 = 121.4 m from the ground