Question

For angles of projection of a projectile(45-theta) and (45+theta) the horizontal ranges described by the projectile are in ratio of???

Answer 1

Answer:

1 : 1

Explanation:

We know for a projectile with initial velocity u and projected at an angle θ, the range is given by

$R=\frac{u^{2}sin2\theta}{g}$

Therefore,

When the angle of projection is (45° - θ)

$R=\frac{u^{2}sin2(45-\theta)}{g}$

⇒ $R=\frac{u^{2}sin(90-2\theta)}{g}$

⇒ $R=\frac{u^{2}cos2\theta}{g}$

Similarly when the angle of projection is (45° + θ)

$R'=\frac{u^{2}sin2(45+\theta)}{g}$

⇒ $R'=\frac{u^{2}sin(90+2\theta)}{g}$

⇒ $R'=\frac{u^{2}cos2\theta}{g}$          (∵ sin (90°+θ) = cosθ)

Thus R = R'

or R/R' = 1

R : R' = 1:1

Answer 2

Answer:

horizontal ranges are in ratio of 1: 1

Explanation:

For angles of projection of a projectile(45-theta) and (45+theta) the horizontal ranges described by the projectile are in ratio of

Case 1 : angles of projection of a projectile(45-θ)

Vertical Velocity = VSin(45-θ)

Time to reach at Peak  = VSin(45-θ)/g

Time for Range = 2VSin(45-θ)/g

Horizontal Velocity = VCos(45-θ)

Horizontal Range = VCos(45-θ)*2VSin(45-θ)/g

= V²2Sin(45-θ)Cos(45-θ)/g

Using Sin2x = 2SinxCosx

= V²Sin(90-2θ)/g

Using Sin(90-x) = Cosx

= V²Cos2θ/g

Case 2 : angles of projection of a projectile(45+θ)

Vertical Velocity = VSin(45+θ)

Time to reach at Peak  = VSin(45+θ)/g

Time for Range = 2VSin(45+θ)/g

Horizontal Velocity = VCos(45+θ)

Horizontal Range = VCos(45+θ)*2VSin(45+θ)/g

= V²2Sin(45+θ)Cos(45+θ)/g

Using Sin2x = 2SinxCosx

= V²Sin(90+2θ)/g

Using Sin(90+x) = Cosx

= V²Cos2θ/g

Horizontal Range in Both cases V²Cos2θ/g

so horizontal ranges are in ratio of 1: 1