when a particle is thrown horizontally ,the resultant velocity of the projectile at any time t is given by
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when a particle is thrown horizontally there is no change in its horizontal velocity (with which it is projected ). So let this horizontal velocity(say u) alone for some time.
further, you will be at some height while projecting the particle hence considering its vertical motion 1D motion we have,
initial velocity = 0, acceleration=10 m/s^2,
then vertical velocity(v) gained by the particle after time t can be given as ,
v= 10t.
okay now let we combine both horizontal and vertical motion ,
velocity is a vector quantity . the angle between u and v is 90° (as one is horizontal other is vertical)
hence the resultant velocity ( velocity of particle at any time is):
=[u^2+(10t)^2]^(1/2)
further, you will be at some height while projecting the particle hence considering its vertical motion 1D motion we have,
initial velocity = 0, acceleration=10 m/s^2,
then vertical velocity(v) gained by the particle after time t can be given as ,
v= 10t.
okay now let we combine both horizontal and vertical motion ,
velocity is a vector quantity . the angle between u and v is 90° (as one is horizontal other is vertical)
hence the resultant velocity ( velocity of particle at any time is):
=[u^2+(10t)^2]^(1/2)
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