Question

A stone is thrown vertically upward whose velocity time

grapgh as shown in figure then find the maximum height reached

by stone , net displacement and total distance covered by it.​

Answer 1

According to the equation of the motion under gravity

v  

2

−u  

2

=2gs

u=initial velocity of the stone=40m/s

v= Final velocity of the stone=0m/s

Let h be the maximum height attained by the stone

Therefore,  

0  

2

−40  

2

=2(−10)h

h=(40×40)/20=80

Therefore, total distance covered by the stone during its upward and  downward journey=80+80=160m

Net displacement during  its upward and downward journey=80+(−80)=0