Question

an object is thrown in vertical upward direction with velocity‘ u ’then the maximum height attained main object would be......... .​

Answer 1

Given :

• The velocity with which the object is thrown in vertical upward direction is 'u'

To find :

• The Maximum height attained by the object

Solution :

When a body/object is thrown vertically upwards the velocity(final velocity) at highest point will be zero.

And the acceleration experienced by the object/body is nothing but the acceleration due to gravity which is equal to -g. (Negative sign indicates that the acceleration is opposite direction of motion)

Initial velocity of the object/body is the velocity with which it is thrown . So , In this case initial velocity of the object will be 'u'.

Now we have ,

• u = u
• v = 0
• a = -g
• s = h (height attained in this case)

From Second equation of motion ,

$\star \: { \boxed{ \sf{ \purple{v {}^{2} - u {}^{2} = 2as }}}}$

Where ,

• v is Final velocity
• u is initial velocity
• a is acceleration
• s is height attained (in this case)

Substituting the values we have ,

$: \implies \sf \: (0) {}^{2} - (u) {}^{2} = 2( - g)(h) \\ \\ : \implies \sf \: 0 - u {}^{2} = - 2gh \\ \\ : \implies \sf \: - u {}^{2} = - 2gh \\ \\ : \implies \sf \not{ - }u {}^{2} = \not{ - }2gh \\ \\ : \implies \sf \: {u}^{2} = 2gh \\ \\ : \implies {\boxed{ \sf {\pink {\: h = \frac{ {u}^{2} }{2g} }}}}$

Hence , The Maximum height attained by the object is $\sf{\dfrac{u^2}{2g}}$

Answer 2

Answer:

Given :

The velocity with which the object is thrown in vertical upward direction is 'u'

To find :

The Maximum height attained by the object

Solution :

When a body/object is thrown vertically upwards the velocity(final velocity) at highest point will be zero.

And the acceleration experienced by the object/body is nothing but the acceleration due to gravity which is equal to -g. (Negative sign indicates that the acceleration is opposite direction of motion)

Initial velocity of the object/body is the velocity with which it is thrown . So , In this case initial velocity of the object will be 'u'.

Now we have ,

u = u

v = 0

a = -g

s = h (height attained in this case)

From Second equation of motion ,

\star \: { \boxed{ \sf{ \purple{v {}^{2} - u {}^{2} = 2as }}}}⋆

v

2

−u

2

=2as

Where ,

v is Final velocity

u is initial velocity

a is acceleration

s is height attained (in this case)

Substituting the values we have ,

\begin{gathered}: \implies \sf \: (0) {}^{2} - (u) {}^{2} = 2( - g)(h) \\ \\ : \implies \sf \: 0 - u {}^{2} = - 2gh \\ \\ : \implies \sf \: - u {}^{2} = - 2gh \\ \\ : \implies \sf \not{ - }u {}^{2} = \not{ - }2gh \\ \\ : \implies \sf \: {u}^{2} = 2gh \\ \\ : \implies {\boxed{ \sf {\pink {\: h = \frac{ {u}^{2} }{2g} }}}}\end{gathered}

:⟹(0)

2

−(u)

2

=2(−g)(h)

:⟹0−u

2

=−2gh

:⟹−u

2

=−2gh

:⟹



−u

2

=



−2gh

:⟹u

2

=2gh

:⟹

h=

2g

u

2

Hence , The Maximum height attained by the object is \sf{\dfrac{u^2}{2g}}

2g

u

2