Question

A ball is projected with a speed of 10 metre per second the two angle of a projection of the range is 5 metre​

Answer 1

Answer:

Keep the following two parametric equations for position in mind, and you’ll solve every projectile motion problem.

Horizontal vector, X(t) = Vo t Cos A

Vertical vector, Y(t) = Vo t Sin A - 1/2 g t^2

We are given: Vo = 20 m / s. To make the calculations easier, let’s assume g = 10 m / s^2. We don’t yet know t-final or angle A.

Range is 10 m. Thus, X(t-final) = 10, which = 20 * t-final * Cos A.

The range, i.e., final distance occurs when Y(t) = 0.

Y(t-final) = 20 * t-final * Sin A - 1/2 * (10) * (t-final)^2 = 0

20 * Sin A - 5 * t-final = 0 (divided by t-final, which is >0)

4 * Sin A = t-final.

We are given that X(t-final) = 10, which = 20 * (4 * Sin A) * Cos A

10 / 20 = 4 * Sin A Cos A

10 / 40 = 2 * Sin A Cos A

1/4 = Sin (2A)

1/2 * Arcsin (1/4) = A

Recall that Sin (B) = Sin (pi - B). Apply this relationship, and you will have derived the two angles that solve the problem.

That is:

1/4 = Sin (pi - 2A)

Arcsin (1/4) = pi - 2A

A = 1/2 * (pi - Arcsin (1/4))

Explanation:

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Answer 2

Answer:

$5 = {10}^{2} \sin(2x) \div 10 \\ \sin(2x ) = 0.5 \\ 2x = {30}^{o} \\ x = {15}^{o}$