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a ball of mass m kg is thrown vertically upwards. Another ball of mass 2m is thrown at angle θ with the vertical. Both stays in air for the same period . the heights reached by them are in the ratio of (a) 2:1 (b) 1:! © 1:cos θ (d) 1:sinθ

a ball of mass m kg is thrown vertically upwards. Another ball of mass 2m is thrown at angle θ with the vertical. Both stays in air for the same period . the heights reached by them are in the ratio of 
 
(a) 2:1
(b) 1:!
© 1:cosθ 
(d) 1:sinθ 
 

Grade:8

1 Answers

Arun
25750 Points
6 years ago
height obtained by an object thrown up with u velocity (h) = u^2/2g
 
Hence, ratio of the heights (h1/h2) = u1^2/u2^2
Also, since they remain in the air for same duration. 
So, using first law of motion v = u + at
v = 0, t = -u/a, Now here, since the a is same as equal to g and t is same, so u needs to be same. 

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