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A ball is droped from a high rise platform t=0 starting from rest. After 6s another ball is thrown downwards from the same platform with a speed v . The two ball meet at t=18s . What is the value of v ?      1)74m/s          2)64m/s .  3)84m/s.    4)94m/s
Asked by Kajalmandal64 | 11 May, 2018, 14:53: PM
answered-by-expert Expert Answer
First ball starting from rest has travelled a distance S1 in 18 seconds. 
 
S1 = (1/2)×9.8×18×18 = 1587.6 m
 
Second ball thrown with initial velocity V,  has travelled a distance S2 in 12 seconds;
 
S2 = V×12+(1/2)×9.8×12×12= (12V+705.6) m;
 
since both balls meet at t=18s, then S1=S2;
hence 12V+705.6 = 1587.6; solving for V, we get V = 73.5 m/s 
Answered by Thiyagarajan K | 11 May, 2018, 15:43: PM
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