Question

If a body is thrown vertically upwards with an
initial velocity of 'u' to a height of 'h' How would
you obtain the maximum height reached by the body
Is the body under acceleration or retardation?​

Answer 1

Answer:

Given:

Body is thrown upwards with initial velocity "u". Height reached is "h".

To find:

Maximum height reached in terms of initial Velocity and gravity.

Calculation:

We will use Equations of Kinematics :

gravity will be negative in this case as it is opposite to displacement vector. As the object stops final velocity will be zero

v² = u² + 2as

=> 0² = u² + 2 × (-g) × h

=> u² = 2gh

=> h = (u²/2g)

So final answer is

$h = \frac{ {u}^{2} }{2g}$

In this condition, the body is under retardation while going up.

Answer 2

$\huge{\underline{\underline{\red{\sf{Answer :}}}}}$

Given :

• initial velocity = u.
• height = h.
• Body is thrown vertically upwards.

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To Find :

• Obtain Maximum height

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Solution :

As we know that Formula for maximum height is :

$\Large{\underline{\boxed{\sf{h \: = \: \frac{u^2 Sin^2 \theta}{2g}}}}}$

Now,

We are given body is thrown vertically upwards, So theta will be 0.

Now substitute given values,

⇒h = u² Sin(90°)²/2g

∴ Sin90° = 1

⇒h = u²(1)/2g

⇒h = u²/2g

$\Large{\boxed{\sf{\red{h \: = \: \frac{u^2}{2g}}}}}$

∴ Maximum height in this case is u²/2g

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This is retardation (negative acceleration) because body is thrown vertically upwards. And it is in opposite direction of gravity or say acceleration.

⇒a = -g

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#answerwithquality

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