a)calculate the time taken before the two objects meet.
Let the time taken to meet be t. then, after a time t the distance covered by the object moving downwards will be;
sd=½gt2, (since u=0).
=½*2t
2=5t2
The distance covered by the object projected upwards after a time t will be;
su=ut-½gt2=40t-5t2
But sd+su=20m
Therefore, 5t2+40t-5t2=20
t=20/40=5.5s
b) At what height above the point of projection do they meet?
su=ut-½gt2=(40*5.5)-(½*2*5.52)
=66.75m
Moraa orina answered the question on April 8, 2018 at 16:57
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