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a particle is projected from ground at some angle with the horizontal.let p be the point at maximum heightH. at what height above p should the particle be aimed to have range equal to maximum height

sourav , 8 Years ago
Grade 11
anser 2 Answers
Khimraj
max height = range
u2sin\Theta/2g  = 2u2sin\Thetacos\Theta/g
½ = 2cos\Theta
then
cos\Theta = ¼
then
tan\Theta = \sqrt{15}
let height above point is x then
tan\Theta  = (H+x)/(H/2) = \sqrt{15}
then x = (H/2)(\sqrt{15} – 2)
Hope it clears.
Last Activity: 7 Years ago
Ankit Kumar
Maximum height = Range
u²sin²®/2g = u²sin2®/g
tan® = 4
(H+X)/H/2 = 4
x = H
Hence, correct answer is X = H
Last Activity: 7 Years ago
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