Question

projectile is projected horizontally with a velocity U show that its trajectory is parabolic . and obtain expression gor time of flight, horizontal range,max height reached,and velocity at any instant​

Answer 1

$\sf\pink{Correct\: Question - }$

An object is projected horizontally with a velocity u. Show that it's trajectory it parabolic and obtain expression for time of flight , horizontal range, max height reached and velocity at any instant.

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$\sf\red{Answer - }$

in x direction -

$\bf u_x = u cos\theta$

$\bf a_x = 0$

in y direction -

$\bf u_y = -u sin\theta$

$\bf a_y = g$

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By using the 2nd equation of motion in x and y direction -

$\bf x = u_x t = u cos\theta t \longrightarrow(1)$

$\bf -y = -u_y t + 1/2 a_y t^2 = -usin\theta(t) + 1/2 gt^2 \longrightarrow (2)$

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From equation (1)

$\bf t = \frac{x}{ \cos\theta}$

Substituting the value in equation (2)

$\bf -y = -u sin\theta( \frac{x}{ucos\theta} )+ \frac{1}{2} g (\frac{x}{ucos\theta} )^{2}$

$\bf y = xtan\theta - \frac{1}{2} \frac{ {x}^{2} }{ {u}^{2} {cos}^{2}\theta }$

As the equation is of $\bf y = ax^2 + bx + c$

So the path is parabolic.

Hence Proved.

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$\sf\pink{Time \:of \:flight - }$

$\bf S = u_y t + 1/2 g t^2$

$\bf 0 = -usin\theta(T) +1/2 g(T)^2$

$\boxed{\bf\red{T = \dfrac{2usin\theta}{g}}}$

where

$\longrightarrow$T is the time of flight.

$\longrightarrow$u is initial velocity.

$\longrightarrow$$\theta$ is the angle of projection.

$\longrightarrow$g is acceleration due to gravity.

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$\sf\pink{Horizontal \:Range - }$

$\bf R = u_x \times T$

$\bf R = u cos\theta \times \frac{2usin\theta}{ g }$

$\bf R = \frac{u^22sin\theta cos\theta}{g}$

$\boxed{\bf\red{R =\dfrac {u^2 sin2\theta}{g}}}$

where -

$\longrightarrow$R is the horizontal range.

$\longrightarrow$u is initial velocity.

$\longrightarrow$$\theta$ is the angle of projection.

$\longrightarrow$g is acceleration due to gravity.

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$\sf\pink{Maximum \:height - }$

Using 3rd equation in y direction -

$\bf V_y^2 = u_y^2 + 2gh$

For max. height $\bf V_y = 0$, $\bf h = H$

$\bf 0 = (-usin\theta)^2 + 2g(-H)$

$\boxed{\bf\red{H = \dfrac{ {u}^{2} {sin}^{2}\theta}{2g}}}$

where -

$\longrightarrow$H is the max. height

$\longrightarrow$u is initial velocity.

$\longrightarrow$$\theta$ is the angle of projection.

$\longrightarrow$g is acceleration due to gravity.

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$\sf\pink{Velocity \:at \:any\: instant - }$

$\bf V_x = u_x = ucos\theta$

$\bf V_y = u_y + at = -usin\theta$

$\bf \vec{V} = V_x \hat{\imath} + V_y \hat{\jmath}$

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Answer 2

Explanation:

An object is projected horizontally with a velocity u. Show that it's trajectory it parabolic and obtain expression for time of flight , horizontal range, max height reached and velocity at any instant.

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Answer:

in x direction -

u_x = u cos\thetau

x

=ucosθ

a_x = 0a

x

=0

in y direction -

u_y = -u sin\thetau

y

=−usinθ

a_y = ga

y

=g

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By using the 2nd equation of motion in x and y direction -

x = u_x t = u cos\theta t \longrightarrow(1)x=u

x

t=ucosθt⟶(1)

-y = -u_y t + 1/2 a_y t^2 = -usin\theta(t) + 1/2 gt^2 \longrightarrow (2)−y=−u

y

t+1/2a

y

t

2

=−usinθ(t)+1/2gt

2

⟶(2)

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From equation (1)

t = \frac{x}{ \cos\theta}t=

cosθ

x

Substituting the value in equation (2)

-y = -u sin\theta( \{x}{ucos\theta} )+ 1/2 g (\frac{x}{ucos\theta} )^{2}−y=−usinθ(

ucosθ

x

)+

2

1

g(

ucosθ

x

)

2

y = xtan\theta - \frac{1}{2} \frac{ {x}^{2} }{ {u}^{2} {cos}^{2}\theta }y=xtanθ−

2

1

u

2

cos

2

θ

x

2

As the equation is of y = ax^2 + bx + cy=ax

2

+bx+c

So the path is parabolic.

Hence Proved.

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Time of flight :

S = u_y t + 1/2 g t^2S=u

y

t+1/2gt

2

0 = -usin\theta(T) +1/2 g(T)^20=−usinθ(T)+1/2g(T)

2

T=

g

2usinθ

where

\longrightarrow⟶ T is the time of flight.

\longrightarrow⟶ u is initial velocity.

\longrightarrow⟶ \thetaθ is the angle of projection.

\longrightarrow⟶ g is acceleration due to gravity.

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Horizontal Range :

R = u_x \times TR=u

x

×T

R = u cos\theta \times \frac{2usin\theta}{ g }R=ucosθ×

g

2usinθ

R =

g

u

2

2sinθcosθ

R= u^2

g

u

2

sin2θ

where -

\longrightarrow⟶ R is the horizontal range.

\longrightarrow⟶ u is initial velocity.

\longrightarrow⟶ \thetaθ is the angle of projection.

\longrightarrow⟶ g is acceleration due to gravity.

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Maximum Height :

Using 3rd equation in y direction -

\bf V_y^2 = u_y^2 + 2ghV

y

2

=u

y

2

+2gh

For max. height V_y = 0V

y

=0 , h = Hh=H

0 = (-usin\theta)^2 + 2g(-H)0=(−usinθ)

2

+2g(−H)

H=

2g

u

2

sin

2

θ

where -

\longrightarrow⟶ H is the max. height

\longrightarrow⟶ u is initial velocity.

\longrightarrow⟶ \thetaθ is the angle of projection.

\longrightarrow⟶ g is acceleration due to gravity.

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Velocity at any instant

V_x = u_x = ucos\thetaV

x

=u

x

=ucosθ

V_y = u_y + at = -usin\thetaV

y

=u

y

+at=−usinθ

V

=V

x

ı

^

+V

y

ȷ

^

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Hope Its Help You.