Question

Two particles are projected with same initial velocity such that one makes an angle P with horizontal while other makes an angle P with vertical.If their common range is R then product of their time of flight is directly proportional to

Answer 1

first particle makes an angle p with horizontal and second p with vertical , its mean second particle makes an angle (90-p) with horizontal .
now both have same Range R
u^2sin2p/g=u^2sin2 (90-p)/g
sin2p=sin2 (90-p)
2p+2p=180
p=45 degree
now
time of flight of 1st projectile (T1)
=2usin45/g
time of flight of 2nd projectile (T2)
=2usin45/g
now
T1 x T2=4u^2sin^2(45)/g^2
=2u^2/g

Answer 2

Pfirst particle makes an angle p with horizontal and second p with vertical , its mean second particle makes an angle (90-p) with horizontal .

now both have same Range R

u^2sin2p/g=u^2sin2 (90-p)/g

sin2p=sin2 (90-p)

2p+2p=180

p=45 degree

now

time of flight of 1st projectile (T1)

=2usin45/g

time of flight of 2nd projectile (T2)

=2usin45/g

now

T1 x T2=4u^2sin^2(45)/g^2

=2u^2/g

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