A particle is projected on to an incline plane of inclination 300 as shown in fig. (g=10m/s2) then:

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the time of flight of the projectile along the inclined plane is 1.63s

the time of flight of the projectile along the inclined plane is 3s

the time of flight of the projectile along the inclined plane is 6.31s

the time of flight of the projectile along the inclined plane is 5s

answer is A.

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Detailed Solution

time of flight=T=2usinα−βgcosβ T=(2)(202)sin(450−300)(10)cos300 T=402sin 1510×32=40×1.414×0.255×1.732=1.63sec

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