Question

A stone is thrown verticly upwards with a velocity of 40ms-1 and returns back taking g=10ms-2 calculating the maximum height reached by the stone and the total distance traveled by the stone​

Answer 1

Explanation:

Initial Velocity u=40

Fianl velocity v=0

Height, s=?

By third equation of motion

v²-u²= 2gs

0−40² =−2×10×s

s=160/2

⇒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.