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Grade: 11


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

5 years ago

Answers : (6)

mohit vashishtha
144 Points
new range will be R+2H
5 years ago
12 Points
How? Will you please explain?
5 years ago
Hasan Naqvi
97 Points
Time taken by projective to reach ground = 2 * (2H/g)1/2
S= ut + ½ at2
= R + ½ * g/2 *2* ((2H/g)1/2)2
=R + g * H/g
=R+ H 
5 years ago
Nicho priyatham
625 Points
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
5 years ago
kritika sharma
11 Points
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.
2 years ago
31 Points
2 years ago
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