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A particle is projected with a velocity "V", so that its range on horizontal plane is twice the greatest height attained. If "g" is acceleration due to grave then its range is :---........ (a) 4V2/5g.... (b) 4g/5V2..... (c) 4V3/5g2..... (d) 4V/5g2...

A particle is projected with a velocity "V", so that its range on horizontal plane is twice the greatest height attained. If "g" is acceleration due to grave then its range is :---........ (a) 4V2/5g.... (b) 4g/5V2..... (c) 4V3/5g2..... (d) 4V/5g2...

Grade:12th pass

6 Answers

Vikas TU
14149 Points
4 years ago
Range => R = V^2sin2thetha/g
and H = V^2/2g
and given as the relation,
R = 2H
V^2sin2thetha/g = V^2/g
sin2thetha = 1
2thetha = pi/2
or
thetha = pi/4
Therfore the Range becomes => 
 R = V^2/g
Yash Gupta
13 Points
3 years ago
ABOVE ANSWER IS WRONG
karthish
13 Points
3 years ago
H = R tan Ф / 4
R = 2 H
 
So tan Ф = 2(or yu can equate R=2H,  ➡v^2sin2 theta/g =2{v^2sin^2theta/g} ➡ tan theta =2) 
 
sin 2Ф = 2 * 2 / (1+2²)  ➡4/1+4 ➡4/5
 
range = v² sin 2Ф / g  ➡4/5v^2/g
 
Raahul
31 Points
3 years ago
H=R tantheta /4
R=2H
 
SO,TAN theta =2(R=2H) 
V*2sin2theta /g=2(v*2sin*2theta/g)
TAN theta=2
Sin2theta =2*2/(1+2*2)
4/1+4
4/5
 
 
Range =v*2 Sin theta /g
4/5v*2g
Anusha
13 Points
2 years ago
A particle is projected with a velocity "V", so that its range on horizontal plane is twice the greatest height attained.
Kushagra Madhukar
askIITians Faculty 629 Points
one year ago
Dear student,
Please find the answer to your question.
 
H = R tan Ф / 4
R = 2 H
So tan Ф = 2 (or you can equate R=2H,  ➡ v2sin2Ф/g = 2{v2sin2Ф/g} ➡ tanФ =2)
sin 2Ф = 2 * 2 / (1+2²)  ➡4/1+4 ➡4/5
range, R = v² sin 2Ф / g  ➡4/5v2/g
 
Thanks and regards,
Kushagra

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