Question

A projectile can have the same range for two angles of projection . if t1 and t2 are the times of flights in the two cases , then the product of the two times of flights is proportional to[AIEEE 2005,04]
(a) R² (b) 1/R² (c) 1/R (d) R ​

Answer 1

Answer:

Time of flight = 2u sin theta / g

Now, we know that the range of projectile is equal for complementary angles,

Since, Range = u^2 sin2theta / g

Now, Product of both time of flight will be

= (2u sin theta/g)*(2u sin(90-theta)/g)

= 2u^2 (2sin theta cos theta)/g)*1/g

= 2/g*(u^2 sin 2theta/g) = 2R/g

Which means,

Product of time of flights is proportional to R

Answer 2

Explanation: - A projectile can have same range , if angle of projection are complementary of each other i.e ¢ and (90° -¢) . this , in both cases ,

• The time of flights are
• t₁= 2usin¢ /g _________(1)
• and , t₂= 2usin(90° -θ)/g ________(1)
• 2ucosθ/g

From equation (1) and (2) ,we get

• t₁ t₂ = 4u²sin¢ cos¢ /g²
• t₁ t₂ = 2u² sin2¢ / g² = 2/g (u²sin2¢/g)
• t₁t₂ = 2R/g [∵ R = u²sin2θ/g]
• t₁ t₂ ∝ R

Hence , option (d) is correct

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