Question

a stone is thrown vertically upward with an initial velocity of 40 metre per second taking G is equal to 10 M per second find the maximum height reached by the stone what is the net displacement and total distance covered by the stone.
please give me full answer.

Answer 1

Answer: The maximum height is 80 m and the total distance covered by the stone is 160 m and the displacement is zero.

Explanation:

Given that,

Final velocity v = 0

Initial velocity u = 40m/s

We know that,

Using equation of motion

$v^2=u^2+2gh$

$0-40^2= 2\times 10\times h$

The maximum height is

$h = 80\ m$

The  stone will reach at the top and will come down

Therefore, the total distance will be

$s = h_{1}+h_{2}$

$s = 80\ m+80\ m = 160 m$

The net displacement is

$D = h_{1}-h_{2}$

$D = 80\ m-80\ m= 0$

Hence, The maximum height is 80 m and the total distance covered by the stone is 160 m and the displacement is zero.

Answer 2

Explanation:

It is given that,

The initial velocity of the stone, u = 40 m/s

We need to find the maximum height reached by the stone, the net displacement and the total distance covered by the stone.

At maximum height, final velocity, v = 0

Using the equation of kinematics as :

$v^2-u^2=2ad$

d is the maximum height reached

a = -g

$d=\dfrac{u^2}{2g}\\\\d=\dfrac{(40)^2}{2\times 10}\\\\d=80\ m$

As the stone goes up and come down. It means that the distance covered by the stone is 2d i.e. 160 m. For this, the displacement of the stone is 0 as the initial position is equal to the final position.