Kinematics Question 190

Question: A particle is projected up with an initial velocity of $ 80ft/\sec $ . The ball will be at a height of $ 96ft $ from the ground after [MP PMT 1985]

Options:

A) 2.0 and 3.0 sec

B) Only at 3.0 sec

C) Only at 2.0 sec

D) After 1 and 2 sec

Show Answer

Answer:

Correct Answer: A

Solution:

$ h=ut-\frac{1}{2}gt^{2}\Rightarrow 96=80t-\frac{32}{2}t^{2} $
$ \Rightarrow t^{2}-5t+6=0\Rightarrow t=2 $ sec or 3 sec



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