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Solve this: p 30 : A ball is thrown from the top of a tower in vertically upward direction. Velocity at a point 'h' meter below the point of projection is twice of the velocity at a point 'h' meter above the point of projection. Find the maximum height reached by the ball above the top of tower : (C) 13) h (D) (4/3) h

The ball is being thrown upwards from the tower with some speed. Let us consider that at a point h meter up from the point of projection the velocity of the ball be v. Now the ball will attain some maximum height and begin falling downwards. Let the velocity at a point h meter below the point of suspension be v'.

Case 1 :- In case of upward motion .
We use formula , v² = u² + 2aS 
Here, a = -g , S = h 
Then, v² = u² - 2gh ∴ v = √(u² - 2gh ) ---------(1) 

Case 2 :- In case of downward motion , 
We use same formula , v² = u² + 2aS 
Here, a = -g , S = -h 
v'² = u² + 2(-g)(-h) = u² + 2gh 
v' = √(u² + 2gh) -----------(2) 

Now, according to the question , 
velocity of ball at height h meters below the point of suspension = 2 × velocity of ball at height h meters below 
⇒ √(u² + 2gh) = 2√(u² - 2gh) 
Squaring both sides, 
⇒ u² + 2gh = 4(u² - 2gh) 
⇒u² + 2gh = 4u² - 8gh 
⇒ -3u² = - 10gh 
⇒u² = 10gh/3 -------(1) 

When the ball is thrown upwards then at maximum height the velocity of the ball will be zero. 
Let the maximum height that the ball reaches be H 
From v² = u² + 2aS 
Here, v = 0, u² = 10gh/3 , a = -g and S = H
Now, 0 = 10gh/3 - 2gH
⇒ H = 5h/3 

Hence, answer is maximum height , H = 5h/3