Minimum velocity

Classical Mechanics Level 4

$P$ and $Q$ are two points a distance $d$ apart at heights $h$ and $k$ above a given horizontal plane. What is the minimum speed $V$ with which a particle can be projected from the horizontal plane so as to pass through $P$ and $Q$ ?

Clarification: the distance $d$ apart is the slanted, point-to-point, separation

V=\sqrt{g(d+h+k)}

V=\sqrt{g(d+2(h+k))}

V=\sqrt{g(2d+h+k)}

V=\sqrt{2g(d+h+k)}

2 solutions

Lu Chee Ket
Oct 17, 2015

Jafar Badour
Oct 17, 2015

if $d=0 \rightarrow h=k$ so the minimum velocity $V=\sqrt{2gh}$ and then the first and second choices are $V=\sqrt{4gh}$ so they're both wrong.

if $h=0$ and $k=d$ the minimum velocity is $V=\sqrt{2gd}$ and the last choice will be $V=\sqrt{3gh}$ so its wrong that yields that the right choice is the 3rd one.

How would you solve this if it was't a multiple choice question?

Miloje Đukanović - 5 years, 8 months ago

Could you please prove the answer?

Lu Chee Ket - 5 years, 8 months ago

Here you go.

Miloje Đukanović - 5 years, 8 months ago

@Miloje Đukanović – I made mistake by taking g h/ (g h + u^2) as u^2/ (g h + u^2) and got doubted. Anyway, I shall read your proof. Thanks!

Lu Chee Ket - 5 years, 7 months ago

Minimum velocity

2 solutions

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