Question

A ball thrown vertically reaches the throwing point after 6 s. If the time taken to reach the highest point by the ball is $x$ and the highest point (in metres) achieved by the ball from the ground is $y$, what is the value of $(x+y)$? (Take $g=9.8m{s}^{-2}$)

• A. 47.1

Answer 1

The correct option is A 47.1
Final velocity of the ball $v=0$

As the ball falls to the same point after it is thrown up, the total displacement of the ball in 6 s is zero.

From newton's second equation of motion,
$h=ut+\frac{1}{2}g{t}^2$
$⟹h=ut+\frac{1}{2}g{t}^2=0$

Here $0=6u+0.5\times (-9.8)\times 6^2$
$⟹$ initial velocity, $u=$ $29.4m{s}^{-1}$

From newton's first equation of motion ,$v=u+gt$
$\therefore$Time taken to reach the highest point $t=\frac{v-u}{g}=\frac{-29.4}{9.8}=3s$ $=x$.

From newton's third equation of motion,$v^2-u^2=2gs$

$\therefore$ Distance of the highest point from the ground, $s=\frac{v^2-u^2}{2g}=\frac{-(29.4{)}^2}{2\times 9.8}$
$=44.1m$ $=y$

$⟹(x+y)=3+44.1=47.1$