Question

If a body is projected vertically up, its velocity decreases to half of its initial velocity at a
height 'h'above the ground. Then maximum height reached by it is
113h
2) 4h
3) 2h
4) 4h/3​

Answer 1

Answer:

• The Maximum Height (H) reached is 4 h / 3 m.

Given:

1. velocity decreases to half of its initial velocity.
2. Let maximum height be " H ".
3. Let the initial velocity be " u ".

Explanation:

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Case - 1

Firstly we need to find the Initial velocity of the body,

From third Kinematic Equation, we Know

$\dashrightarrow\mathrm{v^2 - u^2 = 2\; a\; s}$

Substituting the values,

We know, Final velocity decreases to half of its initial velocity.

Therefore,

$\dashrightarrow\mathrm{\Bigg(\dfrac{u}{2}\Bigg)^2 - u^2 = 2 \times - \; g \times h \ \ \ \because[ s = h]}$

$\dashrightarrow\mathrm{\dfrac{u^2}{4} - u^2 = 2 \times - \; g \times h}$

$\dashrightarrow\mathrm{\dfrac{u^2 - 4u^2}{4} = 2 \times - \; g \times h}$

$\dashrightarrow\mathrm{\dfrac{ - \; 3u^2}{4} = - \;2 \; g \; h}$

$\dashrightarrow\mathrm{\dfrac{ \cancel{-} \; 3u^2}{4} = \cancel{-} \;2 \; g \; h}$

$\dashrightarrow\mathrm{\dfrac{3u^2}{4} = 2 \; g \; h}$

$\dashrightarrow\mathrm{3 \; u^2 = 4 \times 2 \; g \; h}$

$\dashrightarrow\mathrm{3 \; u^2 = 8 \; g \; h}$

$\dashrightarrow{\underline{\underline{\mathrm{u^2 = \dfrac{8 \; g \; h}{3}}}}}$

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Case - 2

Now, Finding the Maximum height (H)

From third Kinematic Equation, we Know

$\dashrightarrow\mathrm{v^2 - u^2 = 2\; a\; s}$

Substituting the values,

At maximum Height Final velocity (v) is zero.

Therefore,

$\dashrightarrow\mathrm{(0)^2 - u^2 = 2\; a\; s}$

$\dashrightarrow\mathrm{- u^2 = 2\; a\; s}$

Now,

$\dashrightarrow\mathrm{- \dfrac{8 \; g \; h}{3} = 2\times - \; g \times H \ \ \ \because \left[u^2 = \dfrac{8 \; g \;h}{3}\right]}$

$\dashrightarrow\mathrm{- \dfrac{8 \; g \; h}{3} = - \; 2 \; g \; H}$

$\dashrightarrow\mathrm{\cancel{-} \dfrac{8 \; g \; h}{3} = \cancel{-} \; 2 \; g \; H}$

$\dashrightarrow\mathrm{ \dfrac{8 \; g \; h}{3} = \; 2 \; g \; H}$

$\dashrightarrow\mathrm{H = \dfrac{8 \; g \; h}{3 \times 2\; g}}$

$\dashrightarrow\mathrm{H = \dfrac{8 \; g \; h}{6\; g}}$

$\dashrightarrow\mathrm{H = \cancel{\dfrac{8 \; g \; h}{6\; g}}}$

$\dashrightarrow{\underline{\boxed{\red{\rm H = \dfrac{4 \; h}{3} \; meters}}}}$

∴ The Maximum Height (H) reached is 4 h / 3 m.

Note:

• The acceleration due to gravity (g) as taken as negative because the body is moving up i.e. Against Gravity.

• H > h

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