Vector Algebra
Projectile Motion
Grade 12
Question:
<p>If \(t_1\) and \(t_2\) are the times of flight of two particles having the same initial velocity \(u\) and range \(R\) on the horizontal, then \(t_1^2 + t_2^2\) is equal to</p>
<p>\(\dfrac{u^2}{g}\)</p>
<p>\(\dfrac{4u^2}{g^2}\)</p>
<p>\(\dfrac{u^2}{2g}\)</p>
<p>\(1\)</p>
Step-by-Step Solution
Key Concept: For projectile motion with same initial speed u and range R, there are two complementary launch angles (θ and 90°-θ). Use the relationships: t = (2u sin θ)/g and R = (u² sin 2θ)/g to find t₁² + t₂² in terms of R and g.
Step 1: For two angles θ_1 and θ_2 giving same range R with initial velocity u: Since R = (u^2 sin 2θ)/g is same for both, we have sin 2θ_1 = sin 2θ_2 This gives: θ_2 = 90° - θ_1 (complementary angles) Step 2: Time of flight for each angle: t_1 = (2u sin θ_1)/g and t_2 = (2u sin θ_2)/g = (2u sin(90° - θ_1))/g = (2u cos θ_1)/g Step 3: Calculate t_1^2 + t_2^2: t_1^2 + t_2^2 = (4u^2 sin^2 θ_1)/g^2 + (4u^2 cos^2 θ_1)/g^2 t_1^2 + t_2^2 = (4u^2(sin^2 θ_1 + cos^2 θ_1))/g^2 = 4u^2/g^2 Step 4: Express in terms of R: From R = (u^2 sin 2θ_1)/g = (2u^2 sin θ_1 cos θ_1)/g, and using t_1t_2 = (4u^2 sin θ_1 cos θ_1)/g^2 We get: t_1t_2 = 2R/g Therefore: t_1^2 + t_2^2 = 4u^2/g^2 or equivalently 2R/g (depending on given options)
Correct Answer: B