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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
- Angle of projection: q1 > q2
- Time of flight: T1 > T2
- Horizontal range: R1 > R2
- Total energy: U1 > U2
Solution
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`
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