Question

A stone is thrown vertically upward with a velocity of 40m/s and taking g = 10m/s .calculate the maximum height reached total distances coverd by th stone and net displacement... please answer with full solution. :-) (»:<br />

Answer 1

S = ut + (1/2)at²

v = u + at

S = Distance or height in this case

v = final speed

u = initial speed

t = time

a = acceleration in this case = g

at max height v =0

u = 40 m/s given

a = g = - 10m/s (as going upward)

0=40-10t

t = 4 sec

S = ut + (1/2)at²

S = 40*4-(10/2)4²

S = 160 - 80

S = 80 m

Height reached = 80 m

Stone will come back to ground

total distance covered = 80 + 80 = 160 m

Displacement = 0 as it will come back to original position