Question

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A boy throws n balls in 2 seconds at regular time intervals. When the first ball reaches the maximum height he throws the second one vertically up. The maximum height reached by each ball is 1) g/(2 * (n - 1) ^ 2) 2) g/(2n ^ 2) 3) g/(8n ^ 2) 4) ) (2g)/(n ^ 2)

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Answer 1

Question:

• A boy throws n balls in 2 seconds at regular time intervals. When the first ball reaches the maximum height he throws the second one vertically up. The maximum height reached by each ball is

Answer:

• The maximum height reached by the ball is 2g / n² ( option D )

Explanation:

Given that :

• A boy throws n balls in 2 seconds at regular time intervals
• When the first ball reaches the maximum height he throws the second one vertically up

To Find:

• The maximum height reached by each ball is

Formula Used:

$\bigstar \;{\underline {\boxed{\bf{ h = \dfrac{u^2}{2g} }}}}$

Required Solution:

• Using formula to find the maximum height

★ We know that ,

${\pink{\bigstar \;{{\boxed{\bf{ h_{(max)} = \dfrac{u^2_{(square \; of \;velocity )}}{2*g_{(gravitational \; acceleration)}} }}}}}}$  

★ Here ,

• Let us denote the height with h
• Let us denote Velocity with u
• Let us denote Time period with t
• Let us denote Number of ball by n

★ Since we know that ,

• The time taken to throw n number of balls upwards is 2 seconds

$\longrightarrow \sf n \; balls = 2 \; s \\ \\ \\ \longrightarrow \sf {\red{\underline{\underline{ 1 \; balls = \dfrac{2}{n} \;s}}}}$  

★  Henceforth

• The time taken to throw 1 ball upward is 2/n seconds

According to the question:

• The boy throws n number of balls upward with regular time intervals and the by the time he throws the second ball upward the first ball reaches its maximum height

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★ Thereby ,

• we can conclude that the first ball reaches its maximum height at the time interval of 2 / n seconds

★ As we know that ,

$\longrightarrow \sf{\red{ time _{(interval)} = \dfrac{u_{(Velocity)}}{g_{(gravitational \; acceleration)}} }}$  

★ Then we can state that ,

$\longrightarrow \sf \dfrac{u_{(velocity)}}{g_{(gravitational \; acceleration)}} = \dfrac{2}{n_{(balls)}} \\ \\ \\ \longrightarrow \sf{\red{\underline{\underline{ u _{(velocity)}= \dfrac{2g}{n} }}}}$

★ Henceforth ,

• Velocity of each ball equals 2g/n

★ Maximum height attained ,

$\longrightarrow \sf height \; reached \; by \; ball _{(max)} = \dfrac{u^2}{2g} \\ \\ \\ \longrightarrow \sf height \; reached \; by \; ball_{(max)} = \dfrac{1}{2g} * n^2 \\ \\ \\ \longrightarrow \sf height \; reached \; by \; ball_{(max)} = \dfrac{1}{\cancel{2g}} * \dfrac{\cancel{2g}}{n} * \dfrac{2g}{n} \\ \\ \\ \longrightarrow \sf {\red{\underline{\underline{ height \; reached \; by \; ball _{(max)} = \dfrac{2g}{n^2} }}}}$  

★ Therefore ,

• The maximum height attained by each of the ball is 2g/n^2

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