Kinematics Question 631

Question: A cricket ball thrown across a field is at heights h1, and h2 from point of projection at times t1 and t2 respectively after the throw. The ball is caught by a fielder at the same height as that of projection. The time of flight of the ball in this journey it’s

Options:

A) h1t22h2t12h1t2h2t1

B)h1t22+h2t12h1t2+h2t1

C) h1t2h1t2h2t1

D) None

Show Answer

Answer:

Correct Answer: A

Solution:

[a] h1=u sinθt1+12gt12;

h2=u sinθt2+12gt22

So, t1t2=h1+12gt12h2+12gt22

h1t2h2t1=12g(t1t22t12t2)

Time of flight = 2u sinθ

g=h1t22h2t12h1t2h2t1 [Use above eqn. to simplify]



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