Question

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Please refer to the attachment..

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Answer 1

Explanation:

at maximum height final velocity will be zero. putting this condition find initial velocity. fallow this process for finding velocity above ground.

Answer 2

Given:

Maximum height attained by ball, h= 20m

To Find:

a) the initial velocity of the ball

b) the final velocity of the ball  on reaching the ground

c) the total time of journey of the ball

Solution:

We know that,

• When a body is thrown vertically upwards, then velocity of body at highest point is 0

• According to first equation of motion for constant acceleration,

$\purple{\boxed{\bf{v=u+at}}}$

• According to second equation of motion for constant acceleration,

$\orange{\boxed{\bf{s=ut+\dfrac{1}{2}at^{2}}}}$

• According to third equation of motion for constant acceleration,

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

where,

v is final velocity

u is initial velocity

a is acceleration

s is displacement

t is time taken

$\rule{190}{1}$

Reference taken here:

• All displacements, velocities, forces and accelerations acting in upward direction are taken positive.

• All displacements, velocities, forces and accelerations acting in downward direction are taken negative.

$\rule{190}{1}$

We have to consider two cases:

Case-1: When ball is going upward

Let the time taken by ball to reach the highest point be t₁, initial velocity be u, final velocity be v' and s be the displacement of ball

On applying third equation of motion for upward motion of ball, we get

$\longrightarrow\rm{(v')^{2}-u^{2}=2as}$

$\longrightarrow\rm{(0)^{2}-u^{2}=2(-g)(20)}$

$\longrightarrow\rm{-u^{2}=-40g}$

$\longrightarrow\rm{u^{2}=40(10)}$

$\longrightarrow\rm{u^{2}=400}$

$\longrightarrow\rm{u=\sqrt{400}}$

$\longrightarrow\rm\green{u=20\:m/s}$

On applying first equation of motion for upward motion of ball, we get

$\longrightarrow\rm{v'=u+at}$

$\longrightarrow\rm{0=20+(-g)t_{1}}$

$\longrightarrow\rm{0=20-gt_{1}}$

$\longrightarrow\rm{gt_{1}=20}$

$\longrightarrow\rm{10(t_{1})=20}$

$\longrightarrow\rm{t_{1}=\dfrac{20}{10}}$

$\longrightarrow\rm{t_{1}=2\:s}$

$\rule{190}{1}$

Case-2: When ball is going downward

Let the time taken by object to reach the ground be t₂, initial velocity be u', final velocity be v and s be the displacement of the object

On applying third equation of motion for downward motion of ball, we get

$\longrightarrow\rm{v^{2}-(u')^{2}=2as}$

$\longrightarrow\rm{v^{2}-0^{2}=2(-g)(-20)}$

$\longrightarrow\rm{v^{2}=2(-10)(-20)}$

$\longrightarrow\rm{v^{2}=400}$

$\longrightarrow\rm{v=\sqrt{400}}$

$\longrightarrow\rm\green{v=20\:m/s}$

On applying second equation of motion for downward motion of ball, we get

$\longrightarrow\rm{s=u't_{2}+\dfrac{1}{2}a(t_{2})^{2}}$

$\longrightarrow\rm{-20=0(t)+\dfrac{1}{2}(-g)(t_{2})^{2}}$

$\longrightarrow\rm{-20=\dfrac{1}{2}(-10)(t_{2})^{2}}$

$\longrightarrow\rm{-20=-5(t_{2})^{2}}$

$\longrightarrow\rm{-5(t_{2})^{2}=-20}$

$\longrightarrow\rm{(t_{2})^{2}=\dfrac{\cancel{-20}}{\cancel{-5}}}$

$\longrightarrow\rm{(t_{2})^{2}=4}$

$\longrightarrow\rm{t_{2}=\sqrt{4}}$

$\longrightarrow\rm{t_{2}=2\:s}$

$\rule{190}{1}$

Let the total time of journey of the ball be T

So,

$\longrightarrow\rm{T=t_{1}+t_{2}}$

$\longrightarrow\rm{T=2+2}$

$\longrightarrow\rm\green{T=4\ s}$

Hence, initial velocity of ball while going upwards is 20 m/s, the final velocity of ball on reaching the ground is 20 m/s and the total time of journey of the ball is 4 s.