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        if a particle is projected from hight H with velocity v & angle ω, what is its maximum range?
7 years ago

SAGAR SINGH - IIT DELHI
879 Points
										Theta is same as w.$x(t) = \frac{}{} v\cos \left(\theta\right) t$
In the vertical direction
$y(t) = \frac{} {} v\sin \left(\theta\right) t - \frac{1} {2} g t^2$
We are interested in the time when the projectile returns to the same height it originated at, thus
$0 = \frac{} {} v\sin \left(\theta\right) t - \frac{1} {2} g t^2$
$\frac{} {}t = 0$
or
$t = \frac{2 v \sin \theta} {g}$

$x = \frac {2 v^2 \cos \left(\theta\right) \sin \left(\theta\right)} {g}$

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7 years ago
gOlU g3n|[0]uS
42 Points


7 years ago
deeksha sharma
40 Points
										To Golu
what you solved was "If a man can throw till max height H, then with same velocity he can throw maximum about 2H long"
whereas the question is simply demanding for the derivation of max range in oblique projection of projectile.

7 years ago
gOlU g3n|[0]uS
42 Points
										hi deeksha then PLZ tell me wright SOL. and correct my mistake

7 years ago
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