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The horizontal range of a projectile is R and the maximum height attained by it is H. A strong wind now begins to blow in the direction of motion of the projectile, giving it a constant horizontal acceleration= g/2. Under the same conditions of projectile, find the horizontal range of the projectile

AP , 9 Years ago
Grade 11
anser 6 Answers
mohit vashishtha

Last Activity: 9 Years ago

new range will be R+2H
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AP

Last Activity: 9 Years ago

How? Will you please explain?

Hasan Naqvi

Last Activity: 9 Years ago

Time taken by projective to reach ground = 2 * (2H/g)1/2
S= ut + Β½ at2
= R + Β½ * g/2 *2* ((2H/g)1/2)2
=R + g * H/g
=R+ H 

Nicho priyatham

Last Activity: 9 Years ago

R+2H 
we know that in 1st  case  H=(u2sin2a)/2g   R=(u2sin2a)/g
now wright the equation of motion in second case
s=((ucosa)t-(1/2)(g/2)t2 )i-((usina)t-(1/2)gt2 )j
when partical lands on ground coffeicant of j =0 so t=(2usina)/g subatute above equation in cofficant of i to  get new range
so we get (u2sin2a)/g +(u2sin2a)/g
=R+2H
 

kritika sharma

Last Activity: 7 Years ago

so its total time of flight and Hmax will be same T=2Vy/g & Hmax=Vy^2/2g
because of the wind horizontal acc. will be g so range will change
Rnew=VxT+1/2 gT^2 =2VxVy/g + 2Vy^2/g
so, 2VxVy/g is normal range & 2Vy^2/g=4H
so ans. is R+4H.
 

Raahul

Last Activity: 6 Years ago

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