Question

A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking
, find the maximum height reached by the stone. What is the net displacement and the
total distance covered by the stone in 6sec?
Pls help

Answer 1

Explanation:

T=6s

S=ut-1/2gt²

(- )since upward direction

let g=10m/s

then

S=40×6-1/2×10×36

S=240-180

S=60m

net displacement =0m

maximum height

H=u²sin²Ø/2g

H=40²×sin²90/2×10

H=1600×1/2

H=800m