Question

A ball thrown up vertically returns to the thrower after $6$s. Find
$\left(a\right)$ The velocity with which it was thrown up,
$\left(b\right)$ The maximum height it reaches, and
$\left(c\right)$ Its position after $4$s.

Answer 1

The ball returns to the ground after $6$ seconds.

Thus the time taken by the ball to reach to the maximum height $\left(h\right)$ is 3 seconds i.e $t=3$ s

Let the velocity with which it is thrown up be $u$

$\left(a\right)$. For upward motion,

$v=u+at$

$\therefore$ $0=u+(-10)\times 3$

$⟹u=30$ $m/s$

$\left(b\right)$. The maximum height reached by the ball

$h=ut+\frac{1}{2}a{t}^2$

$h=30\times 3+\frac{1}{2}(-10)\times 3^2$

$h=45m$

$\left(c\right)$. After $3$ second, it starts to fall down.

Let the distance by which it fall in $1$ s be $d$

$d=0+\frac{1}{2}a{t}^{\prime 2}$ where $t^{\prime }=1$ s

$d=\frac{1}{2}\times 10\times (1{)}^2$ $=5m$

$\therefore$ Its height above the ground, $h^{\prime }=45-5=40m$

Hence after $4s$, the ball is at a height of $40m$ above the ground.