Question

a stone is thrown vertically upward with an initial velocity of 40 m per second taking G is equals to 10 metre per second square find the maximum height reached by the stone what is the net displacement and the total distance covered by the stone​

Answer 1

Answer:

Explanation:

Initial velocity = 40 m/s

g = 10 m/s^2

Time = u/g = 40/10 = 4 seconds

Maximum height reached by stone = u^2/2g

= 40 * 40/2 * 10

= 1600/20 = 80 m

Distance covered = ut + 1/2 gt^2

= 40 * 4 + 1/2 * 10 * 16

= 160 + 80

= 240 m

Answer 2

Answer:

Explanation: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