1yes :)
A projectile is fired with a speed 10 m/s , at an angle 30° with the horizontal.Find the time after which the velocity is perpendicular to the initial velocity of Projection.Take g as
10 m/s^2.
1.2s
2.2.5 s
3.1s
4.None of these
Answer is None of these. Did you get it ?
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16 Answers
1total time period = 1s. At t=1s, then the angle will be 30..
Also since angle is 30., there will never be such a case as the max angle is 30. while landing and it will not reach 60. which is needed to attain the given condition.
1Now let me explain the entire thing.
U = ucos(θ) i + usin(θ) j
V= ucos(θ) i + (usin(θ) - gt) j
if U _|_ V then U . V = 0
Whence ,
u2 - gtusin(θ) = 0
≡ t = u/(gsin(θ))
But this t < T (time of flight)
u/(gsin(θ)) ≤ 2usin(θ)/g
θ ≥ 45° limiting condition.
Infact , at θ = 45° the landing angle is also 45° and the perpendicularity of velocities are achieved at the point of landing. If θ < 45° this condition will never be achieved.
1u=5√3i+5j
v=(5√3i=5j)-10jt
u.v=0
=>(5√3i+5j)[(5√3+5j)-10jt]=0
solving,
t=2
1how can u get this aditya solved it right
{u2 - gtusin(θ) = 0
≡ t = u/(gsin(θ))
}
This step i didnt understand can any one explain it to me??
3we can also change the co-ordinate axis nd then easily get the answer....
1Its easy...the velocity will b perpendicular only wen it wil complete its projectile....i.e., the time perios = 2usin30/10
which is equal to 1 s .....
1and I like a dumbo went on with the whole calculation.. without focusing on the angle..