Question

• A stone is thrown vertically upward with an initial velocity of
40
40 m/s. Taking g= 10 m/s2, find the maximum height reached
by the stone. What is the net displacement and the total
distance covered by the stone?
​

Answer 1

u=40m/s

g= -10m/s^2

v=0m/s

(v)^2-(u)^2=2gh

-(40)^2= -20h

-1600= -20h

1600= 20h

h=80m

distance= 80+80=160m

as the stone goes and comes down with a same height.

displacement=0m

as the stone's inital and final positions are same.

Answer 2

Answer:

The maximum height is 80 m and the total distance covered by the stone is 160 m and the displacement is zero.