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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...

Hasan Rizwi , 8 Years ago
Grade 12th pass
anser 6 Answers
Vikas TU

Last Activity: 8 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

Last Activity: 6 Years ago

ABOVE ANSWER IS WRONG

karthish

Last Activity: 6 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

Last Activity: 6 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

Last Activity: 6 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

Last Activity: 4 Years 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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