Question

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Answer 1

Explanation:

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Answer 2

Question:

A ball is thrown upward with a velocity of 40 m/s. Find maximum height and time to reach there.

$\rule{190}{2}$

Given:

Initial velocity of ball,u= 40 m/s

To Find:

Maximum height attained by ball and time taken by ball to reach there

Solution:

We know that,

• When a body is thrown vertically upwards, then velocity of body at highest point is 0
• According to third equation of motion for constant acceleration,

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

• According to first equation of motion for constant acceleration,

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

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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.

$\rule{190}{1}$

Here,

Final velocity,v= 0 m/s

Let the maximum height attained by ball be 's'

Now, on applying third equation of motion on ball, we get

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

$\longrightarrow\rm{(0)^{2}-(40)^{2}=2(-g)s}$

$\longrightarrow\rm{-1600=2(-10)s}$

$\longrightarrow\rm{-1600=-20s}$

$\longrightarrow\rm{-20s=-1600}$

$\longrightarrow\rm{s=\dfrac{-1600}{-20}}$

$\longrightarrow\rm\green{s=80\ m}$

$\rule{190}{1}$

Let the time taken by ball to attain maximum height be t

So, on applying first equation of motion on ball, we get

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

$\longrightarrow\rm{0=40+(-g)t}$

$\longrightarrow\rm{0=40-10t}$

$\longrightarrow\rm{10t=40}$

$\longrightarrow\rm{t=\dfrac{40}{10}}$

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

Hence, maximum height attained by ball is 80 m and time taken by ball to reach there is 4 s.