A particle is projected with a velocity u so that its horizontal range is thrice the maximum height attained. what is the horizontal range?
Answers
Answered by
92
R=3H
SO WE KNOWS THAT
H/R =TAN THEATA /4
NOW 3H=R
SO TAN THEATA =4/3
I. E. =53°
SO RANGE U^2SIN2X/G
u^2*2sinxcosx/10
=u^2*2*4*3/25*10
12u^2/125
hope it helped tell me if any query on comment
plz mark brainlist god bless you
SO WE KNOWS THAT
H/R =TAN THEATA /4
NOW 3H=R
SO TAN THEATA =4/3
I. E. =53°
SO RANGE U^2SIN2X/G
u^2*2sinxcosx/10
=u^2*2*4*3/25*10
12u^2/125
hope it helped tell me if any query on comment
plz mark brainlist god bless you
Answered by
55
Answer:
The answer is in the attachment!!
hope it helps...mark as brainliesttt.....!!
Attachments:
Similar questions