Question

at what angle teeta with the horizontal hould a body be projected so that its horizontal rng equal the maximum height its attains​

Answer 1

Correct Question:

At what angle (θ) with the horizontal should a body be projected so that its horizontal range equal the maximum height it attains ?

Answer:

• The angle (θ) it attains is tan⁻¹ ( 4 ).

Explanation:

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From the Range Formula we Know,

⇒ R = u² sin 2θ / g

Where,

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

Now,

⇒ R = u² sin 2θ / g

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From the Range Formula we Know,

⇒ H_{max} = u² sin² θ / 2 g

Where,

• H_{max} Denotes Height.
• u Denotes Initial Velocity.
• θ Denotes Angle.
• g Denotes Acceleration due to gravity.

Now,

⇒ H_{max} = u² sin² θ / 2 g

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According to the Question,

⇒ H_{max} = R

Substituting,

⇒ u² sin² θ / 2 g = u² sin 2θ / g

⇒ sin² θ / 2 = sin 2θ

∵ [ sin 2θ = 2 sinθ cosθ ]

Substituting,

⇒ sin² θ / 2 = 2 sinθ cosθ

⇒ sinθ × sinθ = 2 × 2 sinθ cosθ

⇒ sinθ × sinθ = 4 sinθ cosθ

⇒ sinθ  = 4 cosθ

⇒ tan θ = 4

⇒ θ = tan⁻¹ ( 4 )

⇒ θ = tan⁻¹ ( 4 )

∴ The angle (θ) it attains is tan⁻¹ ( 4 ).

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Answer 2

Question

At what angle ∅ with the horizontal should a body be projected so that the horizontal range is equal to the maximum height the projectile attains

Solution

• Angle should be tan^-1(4)

Given

The horizontal range and maximum height should be equal

To finD

Angle of Projection

Now,

Maximum Height is given as :

• $\sf H_{max} = \dfrac{u^2 sin^2 \theta }{2g} ----------(1)$

Horizontal Range is given as :

• $\sf R = \dfrac{u^2 sin2 \theta}{g} ---------(2)$

Equating relations (1) and (2),

$\sf H = R$

$\dashrightarrow \: \sf \: \dfrac{ \cancel{ {u}^{2}} {sin}^{2} \theta }{2 \cancel{g}} = \dfrac{ \cancel{ {u}^{2}} sin2 \theta }{ \cancel{g}}$

Also, sin2∅ = 2sin∅cos∅

$\dashrightarrow \: \sf \: {sin}^{2} \theta = 4sin \theta \: cos \: \theta \\ \\ \dashrightarrow \ \sf \: \frac{sin \: \theta}{cos \: \theta} = 4 \\ \\ \dashrightarrow \: \sf \: tan \theta = 4 \\ \\ \large{ \dashrightarrow \: \boxed{ \boxed{ \sf \theta = {tan}^{ - 1} (4)}}}$