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

6 years ago

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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

$t = \frac{2 v \sin \theta} {g}$

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

Please feel free to post as many doubts on our discussion forum as you can.

All the best.

6 years ago

6 years ago

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.

6 years ago

hi deeksha then PLZ tell me wright SOL. and correct my mistake

6 years ago

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