Question

A particle is thrown vertically upwards. Its velocity at one-fourth of the maximum height is 20 m/s.

Calculate the maximum height attained by the particle ?​

Answer 1

answer is in image good luck for future

Answer 2

Given :

• Initial velocity at ¼ of max. height = 20 m/s.

• Acceleration due to gravity (g) = 9.8 m/s².

To find :

The Maximum height attained by the body.

Solution :

We know the third Equation of Motion i.e,

$\boxed{\bf{v^{2} = u^{2} \pm 2as}}$

But since it is vertical motion or under gravity , we get the third Equation of Motion as :

$\boxed{:\implies \bf{v^{2} = u^{2} \pm 2gh}}$

Now from the third Equation of Motion , we get the Equation for Maximum height as :

$:\implies \bf{v^{2} = u^{2} \pm 2gh}$

[Note : The value of g is taken as postive if it is falling under gravity and the value is taken as negative when it is moving against Gravity]

$:\implies \bf{v^{2} = u^{2} - 2gh}$

[Note :- At the maximum height the Final Velocity of a body is 0]

$:\implies \bf{0^{2} = u^{2} - 2gh}$

$:\implies \bf{0 = u^{2} - 2gh}$

Now subtracting (u²) , from both the sides , we get :

$:\implies \bf{0 - u^{2} = u^{2} - u^{2} - 2gh}$

$:\implies \bf{- u^{2} = \not{u^{2}} - \not{u^{2}} - 2gh}$

$:\implies \bf{- u^{2} = - 2gh}$

$:\implies \bf{\not{-} u^{2} = \not{-}2gh}$

$:\implies \bf{u^{2} = 2gh}$

By dividing (2g) on both the sides , we get :

$:\implies \bf{\dfrac{u^{2}}{2g} = \dfrac{2gh}{2g}}$

$:\implies \bf{\dfrac{u^{2}}{2g} = \dfrac{\not{2}\not{g}h}{\not{2}\not{g}}}$

$:\implies \bf{\dfrac{u^{2}}{2g} = h}$

$\boxed{\therefore \bf{h_{max.} = \dfrac{u^{2}}{2g}}}$

Hence, the Equation for Maximum height is $\bf{\dfrac{u^{2}}{2g}}$

Now , using the above equation and substituting the values in it, we get :

$:\implies \bf{\dfrac{u^{2}}{2g} = h}$

Since, the initial velocity is given when h is ¼, we will multiply the maximum height by ¼ ,i.e

$:\implies \bf{\dfrac{1}{4} \times h = \dfrac{20^{2}}{2 \times 9.8}}$

$:\implies \bf{\dfrac{1}{4} \times h = \dfrac{400}{19.6}}$

$:\implies \bf{\dfrac{1}{4} \times h = 20.4\:(Approx.)}$

$:\implies \bf{h = 20.4 \times 4}$

$:\implies \bf{h = 81.6}$

$\boxed{\therefore \bf{Maximum\:height\:(h) = 81.6\:m}}$

Hence, the maximum height reached by the body is 81.6 m.