Question

A player throws a ball that reaches to the another
player in 4s. If the height of each player is 1.5m
the maximum height attained by the ball from the
ground level is :
(1) 19.6 m
(2) 21.1 m
(3) 23.6 m
(4) 25.1 m​

Answer 1

Given:

Time taken by ball to reach other player,t= 4s

Height of each player,h= 1.5m

To Find:

Maximum height attained by the ball from the  ground level

Solution:

We know that,

In a projectile motion-

• Time of flight T is given by

$\pink{\underline{\boxed{\bf{T=\dfrac{2u\sin\theta}{g}}}}}$

• Maximum height attained by projectile H is given by

$\purple{\underline{\boxed{\bf{H=\dfrac{u^{2}\sin^{2}\theta}{2g}}}}}$

where,

u is initial velocity

θ is the angle by which projectile is projected or thrown

g is acceleration due to gravity

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Let the time of flight of ball thrown be T

So,

$\longrightarrow\rm{T=\dfrac{2u\sin\theta}{g}}$

$\longrightarrow\rm{4=\dfrac{2u\sin\theta}{g}}$

$\longrightarrow\rm{\dfrac{2u\sin\theta}{g}=4}$

$\longrightarrow\rm{u\sin\theta=\dfrac{4g}{2}}$

$\longrightarrow\rm{u\sin\theta=2g}------(1)$

Let the maximum height attained by ball be H

So,

$\longrightarrow\rm{H=\dfrac{u^{2}\sin^{2}\theta}{2g}}$

$\longrightarrow\rm{H=\dfrac{(u\sin\theta)^{2}}{2g}}$

From (1), we get

$\longrightarrow\rm{H=\dfrac{(2g)^{2}}{2g}}$

$\longrightarrow\rm{H=2g}$

$\longrightarrow\rm{H=2\times9.8}$

$\longrightarrow\rm{H=19.6\ m}$

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Let the maximum height attained by the ball from the  ground level be H'

So,

$\longrightarrow\rm{H'=H+h}$

$\longrightarrow\rm{H'=19.6+1.5}$

$\longrightarrow\rm\green{H'=21.1\ m}$

Hence, the maximum height attained by the ball from the  ground level is 21.1 m.

Answer 2

QUESTION:-

✯.ᴀ ᴘʟᴀʏᴇʀ ᴛʜʀᴏᴡs ᴀ ʙᴀʟʟ ᴛʜᴀᴛ ʀᴇᴀᴄʜᴇs ᴛᴏ ᴛʜᴇ ᴀɴᴏᴛʜᴇʀ ᴘʟᴀʏᴇʀ ɪɴ 4s. ɪғ ᴛʜᴇ ʜᴇɪɢʜᴛ ᴏғ ᴇᴀᴄʜ ᴘʟᴀʏᴇʀ ɪs 1.5ᴍ ᴛʜᴇ ᴍᴀxɪᴍᴜᴍ ʜᴇɪɢʜᴛ ᴀᴛᴛᴀɪɴᴇᴅ ʙʏ ᴛʜᴇ ʙᴀʟʟ ғʀᴏᴍ ᴛʜᴇ ɢʀᴏᴜɴᴅ ʟᴇᴠᴇʟ ?

ANSWER✔

$\Large\underline\bold{GIVEN, }$

$\sf\dashrightarrow time\:taken\:(T)=4sec$

$\sf\dashrightarrow height\:of\:the\:player(H)=1.5m$

$\sf\dashrightarrow initial\:velocity=u$

$\sf\dashrightarrow angle\:by\:which\:ball\:is\:thrown = \theta$

$\sf\dashrightarrow acceleration\:because \:of\:gravity=g$

$\Large\underline\bold{TO\:FIND,}$

$\sf\dashrightarrow maximum\:height\:\:attained\:by\:the\:ball.$

$\Large\underline\bold{SOLUTION,}$

$\sf\therefore TIME\:by\:flight\:(T)= \dfrac{2u\sin\theta}{g}$

$\sf\therefore height\:attained(H)= \dfrac{u^{2}\sin^{2}\theta}{2g}$

NOW,

$\sf\implies{T=\dfrac{2u\sin\theta}{g}}$

$\sf\implies{4=\dfrac{2u\sin\theta}{g}}$

$\sf\implies{4g=2u\sin\theta}$

$\sf\implies{u\sin\theta=\dfrac{4g}{2}}$

$\sf\implies{u\sin\theta=\dfrac{\cancel{4g}}{\cancel{2}}}$

$\sf\implies{u\sin\theta=2g}..........eq^1$

NOW,FOR HEIGHT,

$\sf\implies{H=\dfrac{u^{2}\sin^{2}\theta}{2g}}$

$\sf\implies{H=\dfrac{(u\sin\theta)^{2}}{2g}}$

FROM EQ¹ WE GET,

$\sf\implies{H=\dfrac{(2g)^{2}}{2g}}$

$\sf\implies{H=\dfrac{\cancel{(2g)} \times (2g)}{\cancel{2g}}}$

$\sf\implies{H=2g}$

$\sf\implies{H=2\times9.8}$

$\sf\implies{H=19.6\ m}$

$\sf {\fbox { H=19.6m}}$

$\sf\implies{H'=H+h}$

$\sf\implies{H'=19.6+1.5}$

$\sf\implies{H'=21.1\ m}$

$\large{\boxed{\sf{=21.1m}}}$

$\sf\large\therefore hence,\:option\:2\:is\:correct\:21.1m$

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