an object is dropped from a height of 40m and simultaneously another object is thrown upward from the ground with a speed of 40m/s when and where they will meet
Answers
Answered by
2
In this case t=h/u =40/40 =1sec
Height at where they meet =-ut+1/2gt*2
=-40(1)+1/2(10)(1)*2
=-40+5
=-35 m
Height at where they meet =-ut+1/2gt*2
=-40(1)+1/2(10)(1)*2
=-40+5
=-35 m
Answered by
3
answer of this question is here
Attachments:
Similar questions