a-particle-is-projected-from-a-point-at-a-height-3h-metres-above-a-horizontal-play-ground-the-direction-of-the-projectile-makes-an-angle-with-the-horizontal-through-the-point-of-projection-show-t

Question Number 91946 by Rio Michael last updated on 03/May/20

a particle is projected from a point at a height 3 h metres above a horizontal

play ground . the direction of the projectile makes an angle α with the

horizontal through the point of projection . show that if th greatest

height reached above the point lc projection is h metres, then the horizontal

distance travelled by the particle before striking the plane is 6 h \cot α metres .

Find the vertical and horizontal component of the speed of the particle just

before it hits the ground .

Commented by mr W last updated on 04/May/20

a p p l y m e t h o d s a s i n Q 91655

Commented by Rio Michael last updated on 04/May/20

oh sir am sorry i will fix that next time . and thanks

Commented by Rio Michael last updated on 04/May/20

sir in this question, should i assume

the initial velocity is u, and apply that method ?

It seems it is at a hiegt reason why applying the

parabolic property method {doesn}^{'} t work .

Answered by mr W last updated on 04/May/20

M e t h o d I

Commented by Rio Michael last updated on 04/May/20

sir i have some questions about this method of yours .

- do you assume that the origin is at the vertex

that is at the point h on your diagram ? or at O ?

- also when using your expression \frac{y_{1}}{y_{2}} = {(\frac{x_{1}}{x_{2}})}^{2} how do

you choose your points ? that is your coordinates (x_{1} y_{1}) and

(x_{2}, y_{2}) ?

Commented by mr W last updated on 04/May/20

Commented by mr W last updated on 04/May/20

t h e p a t h o f t h e p a r t i c l e i s A B C D .

B i s t h e h i g h e s t p o i n t . t h e p a r a b o l a

i s s y m m e t r i c a b o u t B . w e c a n i m a g e

t h e p a r t i c l e s t a r t s f r o m O . O A i s

t h e m i r r o r i m a g e o f C D .

\tan α = \frac{2 h}{\frac{A C}{2}}

\Rightarrow \frac{A C}{2} = 2 h \cot α

\frac{3 h + h}{h} = {(\frac{{O B}^{'}}{A^{'} B^{'}})}^{2} = 4

{O B}^{'} = 2 A^{'} B^{'} = A C = 4 h \cot α = B^{'} D

A^{'} D = A^{'} B^{'} + B^{'} D = \frac{A C}{2} + B^{'} D

A^{'} D = 2 h \cot α + 4 h \cot α = 6 h \cot α

\Rightarrow p r o v e d

\tan β = \frac{2 (3 h + h)}{B^{'} D} = \frac{8 h}{4 h \cot α} = 2 \tan α

u_{D y} = \sqrt{2 g (h + 3 h)} = 2 \sqrt{2 g h}

u_{D x} = \frac{v_{D x}}{\tan β} = \frac{2 \sqrt{2 g h}}{2 \tan α} = \cot α \sqrt{2 g h}

Commented by mr W last updated on 04/May/20

Commented by Rio Michael last updated on 04/May/20

thanks again sir

Commented by Tawa11 last updated on 06/Nov/21

Great sir

Answered by mr W last updated on 04/May/20

M e t h o d I I

Commented by mr W last updated on 04/May/20

Commented by mr W last updated on 04/May/20

x = u \cos α t

y = u \sin α t - \frac{1}{2} {g t}^{2}

m a x . h i g h t a t t = t_{1} :

u \sin α - {g t}_{1} = 0 \Rightarrow t_{1} = \frac{u \sin α}{g}

y_{B} = u \sin α t_{1} - \frac{1}{2} {g t}_{1}^{2} = h

(u \sin α - \frac{1}{2} g \times \frac{u \sin α}{g}) \frac{u \sin α}{g} = h

\Rightarrow u \sin α = \sqrt{2 g h} \dots (i)

b a l l h i t s t h e g r o u n d a t t = t_{2} :

y_{D} = u \sin α t_{2} - \frac{1}{2} {g t}_{2}^{2} = - 3 h

{g t}_{2}^{2} - 2 \sqrt{2 g h} t_{2} - 6 h = 0

\Rightarrow t_{2} = 3 \sqrt{\frac{2 h}{g}}

x_{D} = u \cos α t_{2} = 3 \sqrt{\frac{2 h}{g}} u \cos α

\Rightarrow u \cos α = \frac{x_{D}}{3 \sqrt{\frac{2 h}{g}}} \dots (i i)

(i) / (i i) :

\tan α = \frac{6 h}{x_{D}}

\Rightarrow x_{D} = 6 h \cot α \Rightarrow p r o v e d

u_{D x} = u \cos α = \frac{6 h \cot α}{3 \sqrt{\frac{2 h}{g}}} = \cot α \sqrt{2 g h}

u_{D y} = u \sin α - {g t}_{2} = \sqrt{2 g h} - 3 \sqrt{2 g h} = - 2 \sqrt{2 g h}

Commented by mr W last updated on 04/May/20

t h e s a m e r e s u l t a s w i t h m e t h o d I .

Commented by Rio Michael last updated on 04/May/20

thank you so much sir