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.

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written 9.3 years ago by

teamques10 ★ 70k

modified 4.2 years ago by

krithikkm200 • 10

Mumbai University > FE > Sem 1 > Engineering Mechanics

Marks : 05

Years : MAY 2015

engineering mechanics

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written 9.3 years ago by

teamques10 ★ 70k

Given:-

For a vertical motion $(\uparrow ve)$

$t=6s, a=- g=-9.81m/s^2$

To find:-

(i) Maximum Height $(h=y_{max})$

(ii)Initial velocity (u)

Solution:

For motion from point 1-2

$ t=3s $ (By symmetry)

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$v=0 \\ a=-9.81m/s^2 \\ u=?$

Using $v=u+at \\ \therefore 0=u-9.81×3 \\ \therefore u=29.43m/s \Rightarrow Ans $

Using $v^2=u^2+2as \\ \therefore 0=(29.43)^2-2×9.81×h \\ \therefore h=\dfrac {(29.43)^2}{2×9.81} \\ \therefore h=44.145m \Rightarrow Ans$

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