A ball is thrown from the top of a tower in vertically upward direction
Answers
Let A ball is thrown upward direction with speed and the ball reaches h height from the top of tower , where velocity is v And after reaching heighest point , ball falls downward . At horizontal line of top of tower , ball again gains velocity v as shown in figure. Now ball is moving downward and falling h height . Let the velocity of ball h height below from the top of tower is v' .
Case 1 :- in case of upward motion .
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 ,
Use same formula , v² = u² + 2aS
Here, a = -g , S = -h
v'² = u² + 2(-g)(-h) = u² + 2gh
v' = √(u² + 2gh) -----------(2)
Now, A/C to question ,
velocity of ball h height below = 2 × velocity of ball h height below
⇒ √(u² + 2gh) = 2√(u² - 2gh)
Squaring both sides,
⇒ u² + 2gh = 4(u² - 2gh)
⇒u² + 2gh = 4u² - 8gh
⇒ -3u² = - 10gh
⇒u² = 10gh/3 -------(1)
∵ ball is throwing upawrd , at maximum height velocity of ball will be zero.
Let ball reaches Maximum height H .
∴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