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# A particle is thrown vertically upwards with some initial velocity.The ratio of distance covered by the particle in time t/2 and 3t/2 is??

Arun Kumar IIT Delhi
askIITians Faculty 256 Points
7 years ago
Since you haven't given condition on v I'm assuming that till 3t/2 its still going up.
$x_{t/2}=vt/2-1/2g(t/2)^2 \\=>x(3t/2)=v(3t/2)-1/2g(3t/2)^2 \\ratio=(v(3t/2)-1/2g(3t/2)^2)/(vt/2-1/2g(t/2))^2 \\=>ratio=(12v-g(3t)^2)/(4v-gt^2)$

nivetha
39 Points
7 years ago
if t is the total time then the body goes up only for t/2 time..then it starts coming down