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Question Number 169317 by BHOOPENDRA last updated on 28/Apr/22

Commented by BHOOPENDRA last updated on 28/Apr/22

$f i n d t i m e o f f l i g h t a l s o ?$
Commented by mr W last updated on 29/Apr/22

$c o n s i d e r a i r r e s i s t a n c e ?$ $t h e n i t h i n k y o u {c a n}^{'} t s o l v e t h e p r o b l e m .$
Commented by BHOOPENDRA last updated on 29/Apr/22

$y e s$
	Answered by mahdipoor last updated on 28/Apr/22

	$V (t) = (- v + v_{0} c o s θ, v_{0} s i n θ - g t)$ $g e t (v_{0} c o s θ - v = a) a n d (v_{0} s i n θ = b)$ $p i s m e a n i n g p o i n t o f r o c k e t$ $V = \frac{d p}{d t} \Rightarrow p (t) = (a t + x_{0}, b t - \frac{{g t}^{2}}{2} + y_{0})$ $\Rightarrow p (0) = (0, l) \Rightarrow x_{0} = 0, y_{0} = l$ $p (t) = (a t, b t - \frac{{g t}^{2}}{2} + l)$
	Commented by BHOOPENDRA last updated on 28/Apr/22

	$w o n d e r f u l s i r b u t y o u a r e l a c k i n g 2, 3$ $t h i n g s . . . L e t m e w a i t f o r m r . W s i r$ $s o l u t i o n .$
	Commented by mr W last updated on 29/Apr/22

	$f r o m m a h d i p o o r {s i r}^{'} s r e s u l t y o u c a n$ $g e t a n s w e r s t o a l l o t h e r q u e d t i o n s$
	Answered by mr W last updated on 30/Apr/22

	$V_{x} = v_{0} \cos θ - v$ $V_{y} = v_{0} \sin θ - g t$ $x (t) = V_{x} t = (v_{0} \cos θ - v) t$ $y (t) = l + v_{0} \sin θ t - \frac{{g t}^{2}}{2}$ $a t y (t) = l + h : V_{y} = 0$ $v_{0} \sin θ - g t = 0 \Rightarrow t = \frac{v_{0} \sin θ}{g}$ $l + h = y {(t)}_{m a x} = l + (v_{0} \sin θ - \frac{g}{2} \times \frac{v_{0} \sin θ}{g}) \frac{v_{0} \sin θ}{g}$ $\Rightarrow h = \frac{v_{0}^{2} \sin^{2} θ}{2 g}$ $a t x (t) = X : y (t) = 0$ $l + v_{0} \sin θ t - \frac{{g t}^{2}}{2} = 0$ $\Rightarrow t = \frac{v_{0} \sin θ + \sqrt{v_{0}^{2} \sin^{2} θ + 2 g l}}{g}$ $X = x (t) = \frac{(v_{0} \cos θ - v) (v_{0} \sin θ + \sqrt{v_{0}^{2} \sin^{2} θ + 2 g l})}{g}$ $R = \sqrt{X^{2} + l^{2}}$ $t o t a l f l i g h t t i m e T = \frac{v_{0} \sin θ + \sqrt{v_{0}^{2} \sin^{2} θ + 2 g l}}{g}$ $t o t a l d i s t a n c e t r a v e l e d :$ $s = \int_{0}^{T} V d t = \int_{0}^{T} \sqrt{V_{x}^{2} + V_{y}^{2}} d t$ $s = \int_{0}^{T} \sqrt{{(v_{0} \cos θ - v)}^{2} + {(v_{0} \sin θ - g t)}^{2}} d t$ $s = - \frac{1}{g} \int_{0}^{T} \sqrt{{(v_{0} \cos θ - v)}^{2} + {(v_{0} \sin θ - g t)}^{2}} d (v_{0} \sin θ - g t)$ $s = \frac{1}{g} \int_{- \sqrt{v_{0}^{2} \sin^{2} θ + 2 g l}}^{v_{0} \sin θ} \sqrt{{(v_{0} \cos θ - v)}^{2} + u^{2}} d u$ $. . . . . .$
	Commented by BHOOPENDRA last updated on 30/Apr/22

	$w e l l d o n e s i r$
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