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A ball is projected vertically upward with an initial velocity of \mathrm{50 \mathrm{~ms}^{-1} at\: t=0 \mathrm{~s}. At\: t=2 \mathrm{~s}},another ball is projected vertically upward with same \mathrm{At \: t= } ______ s,  second ball will meet the first ball \mathrm{\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)}.

Option: 1

6


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

Ball 1
\mathrm{u_{1}=50 \mathrm{~m} / \mathrm{s} \quad(t=0)}

Ball 2
\mathrm{u_{2}=50 \mathrm{~m} / \mathrm{s} \quad(t=25)}

They will meet when the height from the ground is same for both the ball at a given inatant
\mathrm{S_{1}=u_{1} t+\frac{1}{2} a_{1} t^{2} }
\mathrm{S_{2}=u_{2}(t-2)+\frac{1}{2} \, a_{2}\, (t-2)^{2} }

\mathrm{For \; S_{1}=S_{2}}

\mathrm{u_{1}t+\frac{1}{2}\left ( -9 \right )t^{2}= u_{2}\left ( t-2 \right )+\frac{1}{2}+\left ( -9 \right )\left ( t-2 \right )^{2}}
\mathrm{50t-\frac{1}{2}gt^{2}= 50t-100-\frac{1}{2}g\left ( t^{2}-4t+4 \right )}
\mathrm{0= -100+2gt-2g}
\mathrm{120= 20t}
\mathrm{t= 6s}

 

Posted by

Ramraj Saini

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