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 $4\ s$.

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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 13:22:45