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?

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Asked by prakash.sanyasi | 17th Aug, 2019, 12:22: AM

Expert Answer:

u = 40 m/s 
As the stone is thrown upward the acceleration due to gravity is to be taken negative.
g = - 10 m/s2 
v2 - u2 = 2as
For free fall we can write this equation as,
v2 - u2 =2gh
As the stone reaches the maximum height its final velocity v =0
Thus,
0 - (40)2 = 2× (-10) × h 
- 1600 = -20 × h 
h = 80 m 
So, the maximum height to which the stone can reach is 80 m.
 
The total distance covered by the stone = 80 + 80 = 160 m 
And as th stone comes back to its initial position the displacement of the stone = 0

Answered by Shiwani Sawant | 17th Aug, 2019, 10:55: AM