Question

what are the two angles of progection of a projectile projected with velocity 30 metre per second so that the horizontal range is 45 m
(take g=10 m ^-2)​

Answer 1

Answer:

• 15° and 75°

Steps:

According to the concept of Projectile Motion,

Range of a Projectile  is given by the formula:

$\boxed{\text{Range of the projectile} = \dfrac{ u^2 \times sin 2 \theta}{g}}$

where,

'u' refers to the initial velocity of the projectile, 'θ' refers to the angle of projection, 'g' refers to acceleration due to gravity.

According to the question,

• Range = 45 m
• Initial Velocity (u) = 30 m/s
• g = 10 m/s²

Substituting the values in the formula we get:

$\implies 45 = \dfrac{30 \times 30 \times sin 2\theta}{10}\\\\\\\implies 45 \times 10 = 900 \times sin 2\theta\\\\\\\implies 450 = 900 \times sin 2\theta\\\\\\\implies sin 2\theta = \dfrac{450}{900} = \dfrac{1}{2}\\\\\\\implies sin\:2\theta = sin\: 30$

Equating the angles, we get:

→ 2θ = 30°

→ θ = 30° / 2 = 15°

Hence the First Angle of projection is 15°.

We know that, Angles having same range in a projectile motion are always complementary angles. Hence we get:

→ Second Angle = 90° - θ

→ Second Angle = 90° - 15°

⇒ Second Angle = 75°

Therefore the Second Angle of projection is 75°.

Answer 2

Given :

The horizontal range of projection = 45m

Initial velocity = 30m/s

Accerlation due to gravity = 10m/s²

To find :

The angle of projection.

Solution :

The horizontal range of projection is given by

$\underline{ \boxed{ \color{purple}{ \sf R = \dfrac{u^2 sin2 \theta} {g}}}}$

where,

• R denotes horizontal range
• u denotes Velocity
• θ denotes angle of projection
• g denotes gravity

substituting all the given values in the formula,

$:\implies\sf R = \dfrac{u^2 sin2 \theta} {g}$

$:\implies\sf 45= \dfrac{(30)^2 sin2 \theta} {10}$

$:\implies\sf 450= (900 ) sin2 \theta$

$:\implies\sf sin2 \theta= \dfrac{450} {900}$

$:\implies\sf sin2 \theta= \dfrac{1} {2}$

$:\implies\sf sin2 \theta= sin30 \degree$

$:\implies\sf 2 \theta= 30 \degree \: and \: 150 \degree$

$:\implies\sf \theta= 15 \degree \: and \: 75 \degree$

thus, the angle of projection are 15° and 75°.

Remember !

For the given velocity of projection range is same for complimentary angles of projection that is (θ₁ - θ₂ = 90°)