Question

Two balls are projected from the ground at 40
& 50° with respect to the horizontal with
Speed of projection u1 & u2 respectively. If
they have the same horizontal range, then​

Answer 1

Explanation:

when the two angle of projection are complementary angle to each other that is sum is equal to 90⁰. then for a given speed having same horizontal range.

if horizontal range is same then it implies u1 = u2.

Answer 2

Answer:

The relation between u₁, u₂, θ₁, θ₂ is $\frac{u_1}{u_2} =\sqrt{\frac{\sin{2\theta_2}}{\sin{2\theta_1}} }$.

Explanation:

Let R₁ be range of projectile 1 and R₂ be the range of projectile 2.

Given,

θ₁ = 40°, and θ₂ = 50°

Speed of projection of projectile 1 = u₁

Speed of projection of projectile 2 = u₂

Range of projectile (R₁) = Range of projectile (R²)

To find,

Relation between u₁, u₂, θ₁, θ₂

Calculation,

We know that range of a projectile is given by

   $R =\frac{u^2 \sin(2\theta)}{g}$

Then,

$R_1 = \frac{u_1^2 \sin(2\theta_1)}{g} \\R_2 = \frac{u_2^2 \sin(2\theta_2)}{g}$

$\implies \frac{u_1^2 \sin(2\theta_1)}{g} = \frac{u_2^2 \sin(2\theta_2)}{g}$

$\implies \frac{u_1}{u_2} =\sqrt{\frac{\sin{2\theta_2}}{\sin{2\theta_1}} }$

Therefore, the relation between u₁, u₂, θ₁, θ₂ is $\frac{u_1}{u_2} =\sqrt{\frac{\sin{2\theta_2}}{\sin{2\theta_1}} }$.

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