A stone is thrown vertically upward with an initial velocity of 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?
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★★GOOD NIGHT★★
At the top the velocity of stone becomes zero i,e its final velocity becomes zero.
let the maximum height attained by stone be h .
v² - u² = 2gh
v is final velocity and u is intial velocity
u = 40m/5
v = 0
0 - (40)² = 2× -10 × h
1600/20 = h
h = 80m.
Net displacement = 80 - 80 = 0
Total distance = 80 + 80 = 160m
☺️☺️
At the top the velocity of stone becomes zero i,e its final velocity becomes zero.
let the maximum height attained by stone be h .
v² - u² = 2gh
v is final velocity and u is intial velocity
u = 40m/5
v = 0
0 - (40)² = 2× -10 × h
1600/20 = h
h = 80m.
Net displacement = 80 - 80 = 0
Total distance = 80 + 80 = 160m
☺️☺️
Anonymous:
hlw__ please check displacement again__ it will be zero!! I think ☺
Answered by
0
_/\_Hello mate__here is your answer--
____________________
u = 40 m/s
v = 0 m/s
s = Height of the stone
g = −10 ms^−2 ( upward direction)
Let h be the maximum height attained by the stone.
Using equation of motion under gravity
v^2 − u^2 = 2gs
⇒0^2 − 40^2 = 2(−10)ℎ
⇒ ℎ =40×40/ 20 = 80 m
Therefore, total distance covered by the stone during its upward and downward journey
= 80 + 80 = 160 m
Net displacement during its upward and downward journey
= 80 + (−80) = 0
I hope, this will help you.☺
Thank you______❤
_______________________❤
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