Physics, asked by nainajamra, 1 year ago

please answer fast ?????????????????????????.

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Answered by AmanRajSinghania

here u(y) = 20 m/s
(during projectile motion, u(x) remains const, u(y) changes)
using v(y) = u(y) + a(y) t
v(y) = 20 + (-g) 2
= 20 - 20 = 0 m/s.
so v = v(x) i + v(y) j
here v(x) = u(x) = 15 m/s
v = 15 i + 0 j
v = 15 i m/s.....
hope it is correct..

AmanRajSinghania: wlcm!

nainajamra: hhmm

nainajamra: can u explain

nainajamra: please

AmanRajSinghania: are you in 11?

nainajamra: yess

AmanRajSinghania: ok

AmanRajSinghania: in projectile motion, there is no acc. on the particle in x direction... there is only acc. (g) in y direction. so velocity along x remains const, whereas v along y changes. so at any time its v along x is equal to initial velocity along x. So for the question, we know the x component of velocity at t =2.we have to find v along y.. which can be done by using v=u+at.... i hope u will understand..

nainajamra: ok...thanksss

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