Question

A stone is thrown vertically upward with an initial velocity of 20 m/s. Taking g=10 m/s2, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone? ​

Answer 1

As the stone is thrown vertically upward then its Final velocity becomes Zero.

u = 20 ms-¹

By Third equation of Motion:-

v²-u² = 2gh

0² - (20)² = 2 × (-10) × h

-400 = -20h

h = 400/20

h = 20 m

For Displacement,

S = 20 m because body is going in upward direction and its the net displacement transverse by the body and if body comes back to original point then its displacement becomes 0.

Distance = 20 m

Answer 2

Explanation:

Given

u = 20m/s

g = 10m/s^2

vmax = 0

so

v^2 = u^2 - 2gh

2gh = 400

hmax = 400/20

Hmax = 20m

displacement = distance = 20m( if stone doesn't come back to initial position )

if stone come back to original position then

displacement = 0

distance = 20m