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0 votes
41.7k views
in Physics by (6.2k points)

For a given velocity, a projectile has the same range R for two angles of projection if t1 and t2 are the times of flight in the two cases then

(a) t1t∝ R2

(b)  t1t∝ R

(c)  t1t∝1/R

(d)  t1t∝ 1/R2

2 Answers

+1 vote
by (12.0k points)
selected by
 
Best answer

Correct Option  (b) t1t∝ R

Explanation:

For same range angle of projection should be θ and 90 - θ

So, time of flight t1 = 2u sin θ/g and t2 = 2u sin(90- θ)/g = 2u cosθ/g

By multiplying =   t1t2  = 4u2 sinθ cosθ/g2

+1 vote
by (10.9k points)
Let one angle of projection  be A .So other angle of projection will be (90-A). Let t1 and t2 be the time of flight respectively.

So horizontal range in both cases

R=u^2sin2A/g

Now t1=(2usinA)/g

And t2 = (2usin(90-A))/g

So t1t2=(4u^2sinAcosA)/g^2=(2u^2sin2A/g^2

=>t1t2=2R/g

=>t1t2 proportional to R

Option (b)

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