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
