A stone is thrown vertically upwards and returns to the starting point at the ground in 6 sec. Find out max. height and initial velocity of stone.

429views

written 3.9 years ago by

teamques10 ★ 69k

• modified 3.9 years ago

engineering physics

ADD COMMENT EDIT

1 Answer

0views

written 3.9 years ago by

teamques10 ★ 69k

(i)

Given:

$t=6s\\[2ex]s=0\\[2ex]a=19.8m/s_2\\[2ex]s=ut+\frac{1}{2}at^2\\[2ex]0=u\times 6-4.9\times 36\\[2ex]\therefore u=29.4m/s\uparrow$

(ii)

$v=0;s=H\\[2ex]v^2=u^2+2as\\[2ex]\therefore 0=29.4^2+2(-9.8)\times H\\[2ex]\therefore H=44.1m$

Create a free account to keep reading this post.

and 3 others joined a min ago.

ADD COMMENT EDIT

Please log in to add an answer.