Question

A ball is thrown vertically upwards from the top of the tower velocity at a point H M vertically below the point of projection is twice the downward velocity at the point H M vertically above the point of projection the maximum height reached by the ball above the top of the Tower is​

Answer 1

Answer:

Explanation:

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