Here's the question you clicked on:
haridas_mandal
Minimum velocity of a projectile is 20 m/s when a body is projected at an angle of 60 degree with the horizontal .Find the vel of projection....20 m/s,10 m/s,30 m/s or 40 m/s?
so you want the horizontal velocity?
The question i have posted verbatim from my daughter's weekly test. I am not sure but there must be some conditionalities for 20 m/s to be the minimum velocity ..
and there is no diagram or anything?
No none. Only 4 options as I noted above.
it doesnt even have the word "the" at the start of the sentence? Perhaps she copied the question down wrong?
My mistake it should read "The Minimum.. I have the question paper right in front of me..
and it just gives "find the vel of projection"?
It reads " The Minimum.....with the horizontal. The velocity of projection is 1) 20 m/s 2) 10 m/s 3) 30 m/s 4) 40 m/s "
RIGHTTT thats much better. so the minimum velocity is 20m/s in the horizontal direction. find the absolute velocity. okay so i think it looks like this
|dw:1350148494782:dw| your trying to find the hypotenuse
yes, that gives an answer which is one of your possible answers
Her text book says there is a minimum velocity of projection to reach a specific point .I am not very satisfied taking 20 m/s as the vel in the horizontal direction. It should have been 20 Cos 60 as the horizontal component as it remains constant..is there any other possibility?
wel in this isntance ? = 40 m/s
what is the specific point?
How did you deduce? Pl help as I need to understand & then teach my daughter.
Well I just found the hypotenuse of that triangle \[velocity = hypotenuse = \frac{20 m /s}{\cos (60)} = 40 m/s\]