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.

The ball returns to the ground after 6 seconds.

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

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

(a). For upward motion,               

$\mathrm{v}=\mathrm{u}+\mathrm{at}$

∴     $0=\mathrm{u}+(-10)\times 3$           

$⟹\mathrm{u}=30 \mathrm{m}/\mathrm{s}$

Hence the velocity with which it was thrown up is 30 m/s

(b). The maximum height reached by the ball

$\mathrm{h}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$

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

$\mathrm{h}=45 \mathrm{m}$
 

(c). After 3 second, it starts to fall down. 

Let the distance by which it fall in $1 \mathrm{s}$be 'd'

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

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

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

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