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A ball thrown up vertically returns to the thrower after $6\ s$.Find:
$(a)$. the velocity with which it was thrown up,
$(b)$. the maximum height it reaches, and
$(c)$. its position after $4\ s$.
As given, Time to reach Maximum height,
Time, $t=\frac{6}{2}=3\ s$
Velocity, $v=0$ [at the maximum height]
Gravitational acceleration $g=-9.8\ ms^{-2}$
Since the ball is thrown upwards, the acceleration is negative.
$(i)$. The velocity with which it was thrown up,
Using, $v=u+gt$, we get
$0=u-9.8\times 3$
or, $u=29.4\ ms^{-1}$
Thus, the velocity with which it was thrown up $=29.4\ ms^{-1}$
$(ii)$. Maximum height it reaches
Using, $v^2=u^2+2gh$, we get
$h=\frac{v^2-u^2}{2g}$
$h=\frac{0-29.4\times29.4}{(-2\times9.8)}$
$h=44.1\ m$
Thus, Maximum height it reaches$=44.1\ m$
$(iii)$. Position after $4\ seconds$
$t=4s$
In $3\ s$, the ball reaches the maximum height and in $1\ s$ it falls from the top.
Distance covered in $1\ s$ from maximum height,
$h=ut+\frac{1}{2}gt^2$
$h=0+\frac{1}{2}\times9.8\times1$
$h=4.9\ m$
Updated on: 10-Oct-2022
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