A stone is thrown vertically upwards with a velocity 40 m/s and is caught back. Taking
g =10 m/s2, calculate the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?
Answers
Given :
- Initial velocity = 40 m/s
- Acceleration due to gravity = 10 m/s²
- Final velocity = 0 m/s
To find :
- Maximum height reached by the stone
- Net displacement of the stone and the total distance covered when it falls back to the ground
Solution :
To calculate the maximum height reached by the stone, use the first equation of motion.
First equation of motion →
- v² - u² = 2as
where,
v = Final velocity
u = Initial velocity
a = Acceleration due to gravity
s = distance/displacement
we have,
v = 0 m/s
u = 40 m/s
a = - 10 m/s² (Because when the body travels upward the value becomes negative.)
Let us assume the maximum height reached by the stone as s. Substituting the given values :-
→ (0)² - (40)² = 2 × (- 10) × s
→ 0 - 1600 = - 20 × s
→ - 1600 = - 20 × s
→ - 1600/- 20 = s
→ 80 = s
Therefore,
- The maximum height reached by the stone = 80 m
The body travels 80 m upward and then falls down and covers 80 m again. So, the net displacement of the body is zero. As displacement is the distance from the initial position to the final position. And if the body travels a particular distance and come backs at the same point then the displacement becomes zero but the body has travelled some distance.
Distance covered by the body :-
→ 80 + 80
→ 160
Distance covered by the body = 160 m
Therefore,
- Net displacement of the stone is zero
- Distance covered by the stone = 160 m