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?
Answers
Answer:
Initial Velocity u=40
Fianl velocity v=0
Height, s=?
By third equation of motion
v²-u²=2gs
0−40²=−2×10×s
s= 160/20
⇒s=80m/s
Toatl distance travelled by stone = upward distance + downwars distance =2×s=160m
Total Diaplacement =0, Since the initial and final point is same.
Given :-
- A stone is thrown vertically upward with an initial velocity of 40 m/s.
- Acceleration due to gravity (g) = 10 m/s².
To Find :-
- What is the maximum height reached by the stone.
- What is the net displacement.
- What is the total distance covered by the stone.
Formula Used :-
Third Equation Of Motion Formula :
where,
- v = Final Velocity
- u = Initial Velocity
- g = Acceleration due to gravity
- h = Height
Solution :-
Given :-
- Final Velocity (v) = 0 m/s [as stone was at its maximum height]
- Acceleration due to gravity (g) = 10 m/s²
- Initial Velocity (u) = 40 m/s
According to the question by using the formula we get,
The maximum height reached by the stone is 80 m .
As we know that :
Displacement is the shortest distance between the initial and final velocity position of a moving object in particular direction.
So, the net displacement of the stone during it upward and downward :
The net displacement of the stone is zero.
Total distance covered by the stone is :
The total distance covered by the stone is 160 m .