Question

A projectile leaves the ground with a velocity of 30 m/s at an angle of 60 degrees with the horizontal. The maximum height that it could reach is
34.4 m 30.6 m 22.5 m 24.5 m

Answer 1

Answer:

• The maximum height (H_{max}) reached by the projectile is 34.4 meters.

Given:

1. Initial velocity (u) = 30 m/s.
2. Angle of projection (θ) = 60°
3. Acceleration due to gravity (g) = 9.8 m/s²

Explanation:

$\rule{300}{1.5}$

From the formula we know,

⇒ H_{max} = u² sin² θ / 2 g

Where,

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

⇒ H_{max} = u² sin² θ / 2 g

Substituting the values,

⇒ H_{max} = (30)² × sin² 60 / 2 × 9.8

⇒ H_{max} = 900 × ( √(3) / 2 )² / 19.6

⇒ H_{max} = 900 × 3 / 19.6 × 4

⇒ H_{max} = 2700 / 78.4

⇒ H_{max} = 2700 / 78.4

⇒ H_{max} = 34.4

⇒ H_{max} = 34.4 m.

∴ The maximum height (H_{max}) reached by the projectile is 34.4 meters

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Some relations / Formulas:

• R = u² sin 2 θ / g
• T = 2 u sin θ / g
• H_{max} = g t² / 8
• R tan θ = 4 H_{max}

$\rule{300}{1.5}$

Answer 2

Answer:

34.42 m

Explanation:

Given :

Initial velocity u = 30 m / sec

Angle Ф = 60

We are asked to find Vertical height.

We know :

H = u² sin² Ф / 2 g

Putting values here we get :

H = 30² × ( √ 3 )² / 2 × 9.8 × ( 2² )

H = ( 30 × 30 × 3 ) / ( 2 × 9.8 × 4 ) m

H =  ( 15 × 15 × 1.5 ) / ( 9.8 ) m

H = ( 15 × 15 × 15 ) / ( 98 ) m

H = 3375 / 98 m

H = 34.42 m

Hence we get required answer.