A particle is projected with an intial velocity of u ms^(−1) at an angle α to the ground from a point O on the ground. Given that it clears two walls of hieght h and distances 2h and 4h respectively from O. (a) find the tangent of α (b) the maximum hieght (c) the range and period of the particle (d) show that u^2 = (4/(26)) gh please sir can you help me using the actual equations of projectile motion?


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Question Number 91655 by Rio Michael last updated on 02/May/20

$A particle is projected with an intial velocity of u {ms}^{- 1} at an angle α to$ $the ground from a point O on the ground . Given that it clears$ $two walls of hieght h and distances 2 h and 4 h respectively from O .$ $(a) find the tangent of α$ $(b) the maximum hieght$ $(c) the range and period of the particle$ $(d) show that u^{2} = \frac{4}{26} g h$ $please sir can you help me using the actual equations of projectile motion ?$
Commented by Rio Michael last updated on 02/May/20

$no sir, not clear, i still have difficulties solving$ $the problem using your method .$
Commented by mr W last updated on 02/May/20

$i t h o u g h t a l l i s c l e a r i n Q 89319 .$
Commented by mr W last updated on 03/May/20

$a c t u a l l y t h e r e a r e o n l y t h r e e t h i n g s$ $o n e h a s t o k n o w :$ $1 . t h e t r a c k o f t h e b a l l i s a p a r a b o l a .$ $2 . a p a r a b o l a i s a s y m m e t r i c c u r v e .$ $3 . w i t h r e s p e c t t o t h e v e r t e x o f a$ $p a r a b o l a t h e y - c o o r d i n a t e o f a p o i n t i s$ $p r o p o r t i o n a l t o t h e s q u a r e o f i t s$ $x - c o o r d i n a t e, s e e d i a g r a m, i . e .$ $\frac{y_{2}}{y_{1}} = {(\frac{x_{2}}{x_{1}})}^{2}$ $i t h i n k y o u k n o w a l l t h e s e a l r e a d y . s o$ $w h a t y o u s h o u l d d o i s t o a p p l y t h e s e .$
Commented by mr W last updated on 02/May/20

Commented by Rio Michael last updated on 02/May/20

$thank you sir$
Commented by Rio Michael last updated on 02/May/20

$okay sir mr W i really have problems with this particular projectile$ $problem, but sir please check this out .$ $range = 2 h + 2 h + 2 h = 6 h$ $for maximum height using : \frac{y_{1}}{y_{2}} = {(\frac{x_{1}}{x_{2}})}^{2}$ $\Rightarrow \frac{h_{\max}}{h} = \frac{{(3 h)}^{2}}{h^{2}}$ $\Rightarrow h_{\max} = 9 h sir why {doesn}^{'} t this match your previous answer .$
Commented by mr W last updated on 02/May/20

$y o u s h o u l d a p p l y t h e f o r m u l a i n t h e$ $w a y a s t h e p a r a b o l a i n t h e d i a g r a m!$ $i h a v e s a i d : t o a p p l y t h e f o r m u l a t h e$ $y - c o o r d i n a t e a n d x - c o o r d i n a t e m u s t$ $b e w i t h r e s p e c t t o t h e v e r t e x, t h a t$ $s h o u l d b e c l e a r!$ $i n y o u r c a s e i t i s :$ $\frac{h_{m a x}}{h_{m a x} - h} = {(\frac{3 h}{h})}^{2}$
Commented by mr W last updated on 02/May/20

Commented by mr W last updated on 02/May/20

Commented by Rio Michael last updated on 02/May/20

$oh thank God sir i finally got it thanks$
Commented by mr W last updated on 02/May/20

$n e v e r j u s t a p p l y a f o r m u l a b e f o r e$ $o n e h a s u n d e r s t o o d i t!$
Commented by Rio Michael last updated on 02/May/20

$okay sir thanks$
	Answered by mr W last updated on 03/May/20

	$M e t h o d I (q u i c k a n d s a f e, b u t o n e$ $s h o u l d h a v e d e e p u n d e r s t a n d i n g a b o u t$ $p r o j e c t i l e m o t i o n a n d t h e f e a t u r e s$ $o f p a r a b o l a)$
	Commented by mr W last updated on 03/May/20

	Commented by mr W last updated on 03/May/20

	$t h e t r a c k o f t h e b a l l i s a p a r a b o l a w h i c h$ $i s s y m m e t r i c a b o u t t h e p o s i t i o n w h e r e$ $i t r e a c h e s t h e h i g h e s t p o i n t .$ $s i n c e t h e b a l l r e a c h e s t w o w a l l s w i t h$ $t h e s a m e h i g h t h, t h e h i g h e s t p o i n t$ $(s y m m e t r y p o i n t) m u s t b e e x a c t l y i n t h e$ $m i d d l e o f t h e s e t w o w a l l s, s e e d i a g r a m .$ $t h e t r a c k s e c t i o n D B C i s t h e m i r r o r$ $i m a g e o f O A D a b o u t p o i n t D .$ $r a n g e R = O C = 2 h + 2 h + 2 h = 6 h$ $\frac{h_{m a x}}{h_{m a x} - h} = {(\frac{3 h}{h})}^{2} = 9$ $\Rightarrow h_{m a x} = \frac{9}{8} h$ $\tan θ = \frac{2 h_{m a x}}{3 h} = \frac{2 \times 9 h}{8 \times 3 h} = \frac{3}{4} (θ = α i n q n .)$ $t i m e f r o m O t o D i s t h e s a m e a s t i m e$ $f r o m D t o C = \frac{T}{2} = t i m e w h i c h a b a l l$ $n e e d s f o r a f r e e f a l l f r o m h i g h t h_{m a x}$ $t o t h e g r o u n d .$ $\frac{T}{2} = \sqrt{\frac{2 h_{m a x}}{g}}$ $\Rightarrow p e r i o d T = 2 \sqrt{\frac{2 \times 9 h}{8 g}} = 3 \sqrt{\frac{h}{g}}$ $i n t i m e T t h e b a l l m o v e s a h o r i z o n t a l$ $d i s t a n c e R = 6 h,$ $u \cos θ T = R$ $\Rightarrow u = \frac{6 h}{\cos θ T} = \frac{6 h}{T} \times \sqrt{\tan^{2} θ + 1}$ $\Rightarrow u = \frac{6 h}{3 \sqrt{\frac{h}{g}}} \times \sqrt{{(\frac{3}{4})}^{2} + 1} = \frac{5 \sqrt{g h}}{2}$ $(u^{2} = \frac{4}{26} g h g i v e n i n q u e s t i o n i s w r o n g!)$
	Commented by Rio Michael last updated on 03/May/20

	$wow sir thanks so much again . I {don}^{'} t know why$ $this particular projectile motion problem dealt with$ $my understanding . Guess i have to take a whole course$ $on this again, including parabolic paths .$
	Answered by mr W last updated on 03/May/20

	$M e t h o d I I (w h a t m a y b e t h e m o s t$ $p e o p l e d o)$
	Commented by mr W last updated on 03/May/20

	Commented by mr W last updated on 03/May/20

	$x = u \cos θ t$ $y = u \sin θ t - \frac{1}{2} {g t}^{2}$ $x_{A} = u \cos θ t_{A} = 2 h$ $y_{A} = u \sin θ t_{A} - \frac{1}{2} {g t}_{A}^{2} = h$ $x_{B} = u \cos θ t_{B} = 4 h$ $y_{B} = u \sin θ t_{B} - \frac{1}{2} {g t}_{B}^{2} = h$ $x_{B} - x_{A} = u \cos θ (t_{B} - t_{A}) = 4 h - 2 h = 2 h$ $\Rightarrow t_{B} - t_{A} = \frac{2 h}{u \cos θ}$ $y_{B} - y_{B} = u \sin θ (t_{B} - t_{A}) - \frac{1}{2} g (t_{B}^{2} - t_{A}^{2}) = h - h = 0$ $u \sin θ - \frac{1}{2} g (t_{B} + t_{A}) = 0$ $\Rightarrow t_{B} + t_{A} = \frac{2 u \sin θ}{g}$ $2 t_{A} = \frac{2 u \sin θ}{g} - \frac{2 h}{u \cos θ}$ $\Rightarrow t_{A} = \frac{u \sin θ}{g} - \frac{h}{u \cos θ}$ $u \cos θ (\frac{u \sin θ}{g} - \frac{h}{u \cos θ}) = 2 h$ $\Rightarrow u^{2} \sin θ \cos θ = 3 g h$ $(u \sin θ - \frac{1}{2} {g t}_{A}) t_{A} = h$ $[u \sin θ - \frac{1}{2} g (\frac{u \sin θ}{g} - \frac{h}{u \cos θ})] (\frac{u \sin θ}{g} - \frac{h}{u \cos θ}) = h$ $(u \sin θ + \frac{g h}{u \cos θ}) (u \sin θ - \frac{g h}{u \cos θ}) = 2 g h$ $u^{2} \sin^{2} θ - \frac{g^{2} h^{2}}{u^{2} \cos^{2} θ} = 2 g h$ ${(u^{2} \sin θ \cos θ)}^{2} - g^{2} h^{2} = 2 {g h u}^{2} \cos^{2} θ$ $9 g^{2} h^{2} - g^{2} h^{2} = 2 {g h u}^{2} \cos^{2} θ$ $4 g h = u^{2} \cos^{2} θ$ $\Rightarrow u \cos θ = 2 \sqrt{g h}$ $\Rightarrow u \sin θ = \frac{3 \sqrt{g h}}{2}$ $\Rightarrow \tan θ = \frac{3}{4}$ $h_{m a x} a t t = t_{1} :$ $u \sin θ - {g t}_{1} = 0$ $t_{1} = \frac{u \sin θ}{g} = \frac{3 \sqrt{g h}}{2 g} = \frac{3}{2} \sqrt{\frac{h}{g}}$ $h_{m a x} = u \sin θ t_{1} - \frac{1}{2} {g t}_{1}^{2} = \frac{3 \sqrt{g h}}{2} \times \frac{3}{2} \sqrt{\frac{h}{g}} - \frac{1}{2} g \times \frac{9}{4} \times \frac{h}{g}$ $= \frac{9 h}{4} - \frac{9 h}{8} = \frac{9 h}{8}$ $r a n g e a t t = T (= p e r i o d) :$ $u \sin θ T - \frac{1}{2} {g T}^{2} = 0$ $\Rightarrow T = \frac{2 u \sin θ}{g} = \frac{2}{g} \times \frac{3 \sqrt{g h}}{2} = 3 \sqrt{\frac{h}{g}}$ $r a n g e R = u \cos θ T = 2 \sqrt{g h} \times 3 \sqrt{\frac{h}{g}} = 6 h$ $u^{2} = \frac{3 g h}{\sin θ \cos θ} = \frac{3 g h}{\frac{3}{5} \times \frac{4}{5}} = \frac{25 g h}{4}$
	Commented by mr W last updated on 03/May/20

	$w e g e t e x a c t l y t h e s a m e r e s u l t s a s w i t h$ $m e t h o d I .$
	Commented by Rio Michael last updated on 03/May/20

	$wow sir thank you so much for you effort$ $in making me understand this particular problem$ $you are such a {God}^{'} s send . thank you so much sir .$ $am 100 % sure to solve any projectile problem like this .$
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