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 ?
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
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