Question

A Stone is thrown vertically upwards with a velocity of 40m/s and is caught back. (Taking

g=10ms-2)Calculate the maximum height reached by the stone. What is the net displacement

and total distance covered by the stone?​

Answer 1

Answer:

Initial Velocity u=40

Fianl velocity v=0

Height, s=?

By third equation of motion

v

2

−u

2

=2gs

0−40

2

=−2×10×s

s=

20

160

⇒s=80m/s

Toatl distance travelled by stone = upward distance + downwars distance =2×s=160m

Total Diaplacement =0, Since the initial and final point is same.