A ball thrown up vertically returns to the thrower after 6 s. Find.

Solution:

(a) The ball returns to the thrower in 6 s, thus,

the time for its upward journey = 6 ÷ 2 = 3 s

For the upward motion of ball,

initial velocity u = ? ; final velocity v = 0

( $\because$ Ball comes to rest)

Time, t = 3 s

Acceleration due to gravity, $g=-10 \mathrm{~ms}^{-2}$

[In upward direction, $g$ is taken -ve]

v = u + gt

or 0 = u – 10 × 3, or – u = – 30

or $u=30 \mathrm{~ms}^{-1}$

$\therefore$ Initial velocity of ball is $30 \mathrm{~ms}^{-1}$

(b) $\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{gt}^{2}$

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

or $s=90-45=45 \mathrm{~m}$

$\therefore \quad$ Maximum height reached by ball is $45 \mathrm{~m}$.

(c) For the downward motion of ball,

initial velocity, u = 0 ;

time for downward fall, t = 4 – 3 = 1 s ;

acceleration due to gravity, g = 10 ms–2 ;

distance covered in downward direction,s= ?

$\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{~g} \mathrm{t}^{2}$

$s=0 \times(1)+\frac{1}{2} \times 10 \times(1)^{2}$

$\Rightarrow \mathrm{s}=0+5$

$\Rightarrow \mathrm{s}=5 \mathrm{~m}$

$\therefore$ Position of ball after $4 \mathrm{~s}$ from ground

= 45 – 5 = 40 m.