A stone is thrown vertically upward with an initial velocity of 40m/s. Taking g =10m/s, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?
Answers
QUESTION:-
A stone is thrown vertically upward with an initial velocity of 40m/s. Taking g =10m/s, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?
GIVEN:-
- Initial velocity u = 40m/s
- g = 10 m/s2
- Max height final velocity = 0
TO FIND:-
The maximum height reached by the stone and the net displacement and the total distance covered by the stone.
SOLUTION:-
Let us solve using third equation of motion
- v2 = u2 – 2gs
- 0 = (40)2 – 2 x 10 x s
- s = (40 x 40) / 20
Maximum height s = 80m
Total Distance = s + s = 80 + 80
Total Distance = 160m
Total displacement = 0
ANSWER:-
- Maximum height s = 80m
- Total displacement = 0
Hope it helps :D
QUESTION:-
A stone is thrown vertically upward with an initial velocity of 40m/s. Taking g =10m/s, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?
GIVEN:-
Initial velocity u = 40m/s
g = 10 m/s2
Max height final velocity = 0
TO FIND:-
The maximum height reached by the stone and the net displacement and the total distance covered by the stone.
SOLUTION:-
Let us solve using third equation of motion
v2 = u2 – 2gs
0 = (40)2 – 2 x 10 x s
s = (40 x 40) / 20
Maximum height s = 80m
Total Distance = s + s = 80 + 80
Total Distance = 160m
Total displacement = 0
ANSWER:-
Maximum height s = 80m
Total displacement = 0
Hope it helps :D