# A stone is thrown vertically upward with an initial velocity of 40 ms–1. Taking g = 10 ms–2,

Question.
A stone is thrown vertically upward with an initial velocity of 40 ms–1. Taking g = 10 ms–2, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone ?

Solution:

Initial velocity of stone

$\mathrm{u}=40 \mathrm{~ms}^{-1}$

final velocity of stone

v = 0 ;

acceleration due to gravity

$g=-10 \mathrm{~ms}^{2}$

[For upward direction g is taken –ve] ;

height attained by stone (h) = ?

We, know, $\mathrm{v}^{2}-\mathrm{u}^{2}=2 \mathrm{gh}$ or $(0)^{2}-(40)^{2}=2 \times(-10) \times \mathrm{h}$

or $h=\frac{-1600}{-20}=80 \mathrm{~m}$

$\therefore$ Maximum height attained by stone $=80 \mathrm{~m}$

Net displacement of stone = 0

(because the stone returns to the same point)

Total distance covered by the stone = 2 × height attained = 2 × 80 = 160 m