Question

a ball thrown in upward direction acceleration g=10m/sec from the ground with speed of 200m/sec after 8sec it comes back to the ground find the total distance covered by body and what will be the velocity of the ball at maximum height

Answer 1

$\huge\it\red{\underline{\underline{Answer:-}}}$

$\tt{1. \ The \ total \ distance \ covered \ by \ the \ ball}$

$\tt{is \ 40 \ m}$

$\tt{2. \ The \ velocity \ of \ the \ ball}$

$\tt{will \ be \ zero \ at \ the \ maximum \ height.}$

$\huge\it\orange{Given:}$

$\sf{\leadsto{Acceleration \ due \ to \ gravity (a)=10 \ m \ s^{-2}}}$

$\sf{\leadsto{Initial \ velocity (u)=200 \ m \ s^{-1}}}$

$\sf{\leadsto{Total \ time \ taken \ by \ ball \ to \ go \ up}}$

$\sf{and \ come \ down=8 \ s}$

$\huge\it\pink{To \ find:}$

$\sf{1. \ Maximum \ distance \ cover \ by \ the \ ball.}$

$\sf{2. \ Ball's \ velocity \ at \ maximum \ height.}$

$\huge\it\green{\underline{\underline{Solution:}}}$

$\sf{1.}$

$\sf{We \ know, \ time \ taken \ for \ ball \ to \ go}$

$\sf{upward \ is \ same \ as \ time \ taken \ for \ ball}$

$\sf{to \ go \ downward.}$

$\sf{Hence, \ time \ taken \ to \ go \ upward=\dfrac{8}{2}}$

$\sf{\therefore{Time \ taken \ for \ ball \ to \ go \ up=4 \ s}}$

$\sf{Now, \ for \ Upward \ direction:}$

$\sf{\leadsto{Time(t)=4 \ s,}}$

$\sf{\leadsto{Initial \ velocity (u)=200 \ m \ s^{-1}}}$

$\sf{\leadsto{Final \ velocity (v)=0}}$

$\sf{\leadsto{Acceleration \ due \ to \ gravity=-10 \ m \ s^{-2}}}$

$\sf{According \ to \ the \ third \ equation \ of \ motion}$

$\boxed{\sf{v^{2}=u^{2}+2as}}$

$\sf{\therefore{0=200^{2}+2(-10)s}}$

$\sf{\therefore{-400=-20s}}$

$\sf{\therefore{s=\dfrac{-400}{-20}}}$

$\sf{\therefore{s=20 \ m}}$

$\sf{Total \ distance \ covered=2\times20}$

$\sf{\therefore{Total \ distance \ covered=40 \ m}}$

$\tt\purple{The \ total \ distance \ covered \ by \ ball \ the}$

$\tt\purple{is \ 40 \ m}$

____________________________________

$\sf{2. \ For \ upward \ direction \ always \ the}$

$\sf{final \ velocity \ is \ zero.}$

$\tt\purple{Hence, \ the \ velocity \ of \ the \ ball}$

$\tt\purple{will \ be \ zero \ at \ the \ maximum \ height.}$