Projection Time of Flight Product – Rankers Physics
Topic: Kinematics
Subtopic: Projectile Motion

Projection Time of Flight Product

Two particles are projected with same initial velocity one makes angle \(\theta\) with horizontal while other makes an angle \(\theta\) with vertical. If their common range is R then product of their time of flight is directly proportional to:

(1999)

R
\(R^{2}\)
\(\frac{1}{R}\)
\(R^{0}\)

Solution:

Concept: Time of flight for complementary angles and range formula.
Formula: \(T = (2u sin \alpha) / g\), \(R = (u^2 sin 2\alpha) / g\).
For angles \(\theta\) and \(90° - \theta\), times are \(T_1 = (2u sin \theta) / g\) and \(T_2 = (2u cos \theta) / g\).
Their product \(T_1 T_2 = (4u^2 sin \theta cos \theta) / g^2 = (2u^2 sin 2\theta) / g^2 = (2/g) R\). Thus, \(T_1 T_2 \propto R\).

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