Question
A ball is thrown vertically upwards. It returns 6 s later. Calculate : The greatest height reached by the ball . (Take $g =10 m s ^{-2}$ )
✓
Answer
Total time of journey $=6 s$
$
g =10 m / s ^2
$
Let ' $H$ ' be the greatest height.
Time of ascent, $t=6 / 2=3 s$,
For ascent, initial velocity, $u =0$
Using the second equation of motion,
$
\begin{aligned}
& H=u t+(1 / 2) g t^2 \\
& H=0+(1 / 2)(10)(3)^2 \\
& H=45 m
\end{aligned}
$
$
g =10 m / s ^2
$
Let ' $H$ ' be the greatest height.
Time of ascent, $t=6 / 2=3 s$,
For ascent, initial velocity, $u =0$
Using the second equation of motion,
$
\begin{aligned}
& H=u t+(1 / 2) g t^2 \\
& H=0+(1 / 2)(10)(3)^2 \\
& H=45 m
\end{aligned}
$
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