A stone is thrown vertically upward with an initial velocity of 40 m/so. taking g=10 meters /square .find the maximum height reached by the the stone .what is net displacement and the total distance covered by the stone?
Answers
Answered by
0
h= 80 m
dispacement 0m
total distance will be 80m
dispacement 0m
total distance will be 80m
Answered by
2
hey...
u = 40 m/s
g = 10 m/s^2
height= ?
v= 0
by using
2gs = v^2 - u^2
2× 10 × s = 0^2 - 40^2
20 × s = - 1600
s= 1600/20
s = 80 m.
distance = 80 + 80 = 160( upward + downward direction)
displacement = 0
u = 40 m/s
g = 10 m/s^2
height= ?
v= 0
by using
2gs = v^2 - u^2
2× 10 × s = 0^2 - 40^2
20 × s = - 1600
s= 1600/20
s = 80 m.
distance = 80 + 80 = 160( upward + downward direction)
displacement = 0
Similar questions