A ball is projected vertically upwards with velocity v if another ball in the same manner thrown (i.e. with v and vertically upwards ) after 'n' second n can be ant positive number
1) find time after wich ball will meet (if possible)
2) hieght from projection plane at which they will meet?
1
Expert's answer
2018-09-25T10:22:09-0400
1) For the first ball: h=v(t+n)-(g(t+n)^2)/2 For the second ball: h=vt-(gt^2)/2 So, v(t+n)-(g(t+n)^2)/2=vt-(gt^2)/2 vn=g/2(n^2+2tn) v=g/2(n+2t) t=(2v/g-n)/2=v/g-n/2. 2) h=v(v/g-n/2)-g/2 (v/g-n/2)^2 h=v^2/2g-(gn^2)/8
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