Question

A stone is thrown upward with initial velocity of 40 metre square taken 10 cm square find the maximum height reached by the stone what is the net displacement and the total distance covered by the stone stone

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

mark me as brainliests

Answer 2

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

v² = u²+2gh

0 - 40²= 2×10×h

The maximum height is

h= 80m

The stone will reach at the top and will come down

Therefore, the total distance will be

s = h1 + h2

s = 80m + 80m = 160m

The net displacement is

D = h1 - h2

D = 80m - 80m = 0

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

Explanation:

hope it helps...