Question

Two particles are projected in air with speed vo at angles θ₁ and θ₂ (both acute) to the horizontal, respectively. If the height reached by the first particle is greater than that of the second, then tick the right choices

1. Angle of projection: q₁ > q₂
2. Time of flight: T₁ > T₂
3. Horizontal range: R₁ > R₂
4. Total energy: U₁ > U₂

Answer 1

a and b

Explanation:

According to formula, Max height of a projectile is 

H = `(u^2 sin^2 theta)/(2g)`

a. `H_1 > H_2`

`sin^2 theta_1 > sin^2 theta_2`

`(sin theta_1 + sin theta_2) (sin theta_1 - sin theta_2) > 0`

So, `(sin theta_1 + sin theta_2) > 0` or `(sin theta_1 - sin theta_2) > 0`

So, `θ_1 > θ_2` and θ lies between 0 and 90 degree i.e. acute

b. `T_1/T_2 = sin theta_1/sin theta_2`

`T_1/T_2 = sin theta_1/sin theta_2`

`T_1  sin theta_2 = T_2  sin theta_1`

Since sin θ₁ > sin θ₂

T₁ > T₂ 

Important points about the time of flight: For complementary angles of projection θ and 90° – θ.

a. Ratio of time flight = `T_1/T_2`

= `(2 u  sin  theta/g)/(2u  sin  (90^circ - theta)/g)`

= tan θ

⇒ `T_1/T_2` = tan θ

b. Multiplication of time of flight = `T_1T_2`

= `(2u  sin theta)/g  (2u cos theta)/g`

⇒ `T_1T_2 = (2R)/g`