Question

a stone is thrown vertically upward with an initial velocity of 40 metre per second taking G is equal to 10 metre per second square find the maximum height reached by the stone what is the net displacement and total distance covered by the stone​

Answer 1

Answer:

u=40m/s

a=g=-10m/s^2

v=0m/s

v^2-u^2=2as

-1600=-10s

s=160m

and total distance=160m+160m=320m

and total displacement is 0m because ball is finally return to it's initial position