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
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