Question

A stone is thrown vertically upward with an initial velocity of 40 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

Data:

u=40m/s

v=0m/s

g=-10m/s ( upwards)

s=?

2gs=v^2-u^2

2× -10m/sec× s =0-40^2

-20m/sec×s=-1600m/s

s=-1600m/sec/-20m/sec

s=80 meters

Now, again come backs to initial position

Hence, total distance covered by stone =80(upwards)+80(downwards)=160 meters

Answer 2

A stone is thrown with speed 40 m/s , at highest point velocity of stone = 0 m/sec

use , kinematics equation ,

V² = U² + 2aS

where V = 0

U = 40 m/s

a = -g m/s² = -10 m/s²

0 = (40)² -2 × 10 × S

S = 80 m .

hence maximum hight is reached by stone = 80 m .

now, stone comes back to intial point.

hence

total distance covered by stone = 80 ( in upward motion ) + 80( in downward motion ) = 160 m .