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Q.

A projectile is thrown with velocity v making an angle θ with the horizontal. It just crosses the tops of two poles, each of height h, after 1 s and 3 s respectively. The time of flight of the projectile is

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a

4 s

b

3 s

c

7.8 s 

d

1 s

answer is C.

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Detailed Solution

h=vsinθt−12gt2

or    12gt2−vsinθt+h=0

t1+t2=−−vsinθ12g

or         t1+t2=2vsinθg=T

T=(1+3)s=4s

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