Find the maximum height that an object can achieve if it is thrown in the opposite direction at a speed of 20 m/s
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Explanation :- case 1 :-when body moves upward direction.
initial velocity =u
final velocity =v=0
now, v=u+at
Here, v=0,a=−g [acceleration is downward direction]
0=u−gt⇒t=u/g
Maximum height = h
u
2
=2gh⇒h=u
2
/2g---(1)
Case 2 :- when body moves downward direction.[Falling from maximum height]
initial velocity, u=0
final velocity = v
use formula, s=ut+1/2at
2
here,s=−h,u=0 and a=−g
−h=0−1/2gt
′2
−u
2
/2g=−gt
′2
/2 [from equation(1)]
t
′
=u/g
Hence, total time =t+t
′
=u/g+u/g=2u/g
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