Question

A body is projected at such an angle that horizontal
range is three times the greatest height. The angle
of projection is-
(1) 30° (2) 37° (3) 45° (4) 53°​

Answer 1

Answer:

53°

4. option is correct .

Explanation:

Given :

Horizontal range R = 3 vertical height

We know :

R = u² sin 2 Ф / g

H =  u² sin² Ф /2  g

= > u² sin 2 Ф / g = 3 ( u² sin² Ф / 2 g )

= > sin 2 Ф =  3 / 2 × sin² Ф

We know :

sin 2 Ф = 2 sin Ф cos Ф

= > 2 sin Ф cos Ф = 3 / 2 . sin Ф . sin Ф

= > 4 cos Ф = 3 sin Ф

= > sin Ф / cos Ф = 4 / 3

= > tan Ф = 4 / 3

= > Ф = tan⁻¹ ( 4 / 3 )

= > Ф = 53°

Hence we get answer.

Answer 2

Answer:

• The Angle of projection (θ) of the body is 53 °

Given:

1. Range is three times the Maximum height

Explanation:

$\rule{300}{1.5}$

From the relation we know,

⇒ H_{max} = u² sin² θ / 2 g

Where,

• H_{max} Denotes maximum height.
• u Denotes Initial velocity.
• g Denotes Acceleration due to gravity.
• θ Denotes Angle.

Now,

⇒ H_{max} = u² sin² θ / 2 g

$\rule{300}{1.5}$

$\rule{300}{1.5}$

From the relation we know,

⇒ R = 2 u² sin 2θ / g

Where,

• Range Denotes Range.
• u Denotes Initial velocity.
• g Denotes Acceleration due to gravity.
• θ Denotes Angle.

Now,

⇒ R = u² sin 2θ / g

$\rule{300}{1.5}$

$\rule{300}{1.5}$

According to the question

⇒ R = 3 H_{max}

Substituting the values,

⇒ u² sin 2θ / g = 3 × ( u² sin² θ / 2 g )

⇒ u² sin 2θ / g =   3 u² sin² θ / 2 g

∵ [ sin 2θ = 2 sinθ cosθ ]

⇒ u² 2 sinθ cosθ  / g =   3 u² sin² θ / 2 g

⇒ 2 sinθ cosθ = 3 sin² θ / 2

⇒ 3 sin² θ = 4 sinθ cosθ

⇒ 3 sin θ × sin θ = 4 sinθ cosθ

⇒ 3 sin θ  = 4 cosθ

⇒ sin θ / cos θ = 4 / 3

⇒ tan θ = 4 / 3

⇒ θ = tan⁻¹ ( 4 / 3 )

⇒ θ = 53 °

⇒ θ = 53 °

∴ The Angle of projection (θ) of the body is 53 °.

$\rule{300}{1.5}$