Question

show that the range of a projectile for two angle of projection alpha and beta is same where alpha + beta is equal to 90 degree

Answer 1

Answer:

yes, complementary angle horizontal angal are always equal

Explanation:

α+β=90°

α and β are complementary angle

for the angle α

R1=u2sinthetaα/9

R2=u2sinthetaβ/9

= u2sin2(90°-α)

=u2sin(180°-2α)

= u2sin2α/9

=R1

hence, for complementary angle horizontal angle are always equl

Answer 2

Answer:

The range of a projectile for two angles of projection alpha and beta is the same.

Explanation:

The range of a projectile is given as,

$R=\frac{u^2sin2\theta}{g}$             (1)

Where,

R=range of projectile

u=velocity of projection

θ=angle of projection

g=acceleration due to gravity=10m/s²

As per the question, two projection angles are α and β. And let β be 90°-α.

θ₁=α

θ₂=90°-α

The range for θ₁ is given as,

$R_1=\frac{u^2sin2\alpha}{g}$                (2)

The range for θ₂ is given as,

$R_2=\frac{u^2sin2(90\textdegree-\alpha)}{g}$          (3)

∴ sin2(90°-θ)=sin 2θ    (4)

By using equation (4) in equation (3) we get;

$R_2=\frac{u^2sin2(90\textdegree-\alpha)}{g}=\frac{u^2sin2\alpha}{g}$     (5)

From equations (2) and (5) we get;

R₁=R₂

Thus horizontal range is the same for the angle of projection α and 90°-α.

Hence, α + β=90°.