Question

two particles are projected with same velocity but at angles of projection 35 degrees and 55 degrees then their horizontal ranges are in the ratio ?

Answer 1

Answer:

$\dfrac{R_1}{R_2}=\dfrac{1}{1}$

Explanation:

Angle of projection of ball 1, $\theta_1=35^{\circ}$

Angle of projection of ball 2, $\theta_2=55^{\circ}$

The horizontal range of a projectile is given by :

$R=\dfrac{v^2\ sin2\theta}{g}$

For first ball, $R_1=\dfrac{v^2\ sin2(35)}{g}$............(1)

For second ball, $R_2=\dfrac{v^2\ sin2(55)}{g}$...........(2)

Taking ratios of equation (1) and (2) as :

$\dfrac{R_1}{R_2}=\dfrac{sin(70)}{sin(110)}$

$\dfrac{R_1}{R_2}=\dfrac{1}{1}$

So, the ratio of their horizontal ranges are in the ratio of 1:1. Hence, this is the required solution.

Answer 2

Explanation:

Answer:

\dfrac{R_1}{R_2}=\dfrac{1}{1}

R

2

R

1

=

1

1

Explanation:

Angle of projection of ball 1, \theta_1=35^{\circ}θ

1

=35

∘

Angle of projection of ball 2, \theta_2=55^{\circ}θ

2

=55

∘

The horizontal range of a projectile is given by :

R=\dfrac{v^2\ sin2\theta}{g}R=

g

v

2

sin2θ

For first ball, R_1=\dfrac{v^2\ sin2(35)}{g}R

1

=

g

v

2

sin2(35)

............(1)

For second ball, R_2=\dfrac{v^2\ sin2(55)}{g}R

2

=

g

v

2

sin2(55)

...........(2)

Taking ratios of equation (1) and (2) as :

\dfrac{R_1}{R_2}=\dfrac{sin(70)}{sin(110)}

R

2

R

1

=

sin(110)

sin(70)

\dfrac{R_1}{R_2}=\dfrac{1}{1}

R

2

R

1

=

1

1

So, the ratio of their horizontal ranges are in the ratio of 1:1. Hence, this is the required solution.