solve question no. 133 plzz.
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Given ,
horizontal range =200 m
time of flight = 5 sec
g= 10 m/s^2
we know ,
according to oblique projectile
any body projected with speed v inclined with horizontal by @
then,
x -component of v =vcos@
y -component of v = vsin@
and Range = v^2sin2@/g
=v^2([email protected]@)/g
=2 (vcos@)(vsin@)/g
=2 (x-component of v)(y-component of v)/g
and also
time of flight =2vsin@/g
=2 (y-component of v)/g
time of flight =5 sec given ,
so,
5 =2 (y-component of v)/g
y -component of v= 5g/2
now ,
put this in range formula ,
200 =2 (x-component of v)(5g/2)/g
200=5 (x -component of v)
x -component of v= 40 m/s
hence horizontal speed = x-component of v = 40 m/s
horizontal range =200 m
time of flight = 5 sec
g= 10 m/s^2
we know ,
according to oblique projectile
any body projected with speed v inclined with horizontal by @
then,
x -component of v =vcos@
y -component of v = vsin@
and Range = v^2sin2@/g
=v^2([email protected]@)/g
=2 (vcos@)(vsin@)/g
=2 (x-component of v)(y-component of v)/g
and also
time of flight =2vsin@/g
=2 (y-component of v)/g
time of flight =5 sec given ,
so,
5 =2 (y-component of v)/g
y -component of v= 5g/2
now ,
put this in range formula ,
200 =2 (x-component of v)(5g/2)/g
200=5 (x -component of v)
x -component of v= 40 m/s
hence horizontal speed = x-component of v = 40 m/s
abhi178:
here we understand concept
that, is the concept to be understood.
exchange own knowledge
so objective is not enclude here
Whatever you say.
you understand it s not mean another person also understood
I get the exact method from abhi.....n I find Vishnu's method to be easy for competitive point of view
I mean objective answer not give object through okay
so thnx 2 both of you....n vishnu...... I will become ur follower.......so whenever I will ask any question..... I want you to answer my question this time.... I promise if I find it appropriate I will definitely mark u as brainliest
I also know Vishnu your trick but I apply only my question .
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