Two particles are projected in air with speed vo at angles θ1 and θ2 (both acute) to the horizontal, respectively. If the height reached by the first particle is greater than that of the second, then tick the right choices
(a) Angle of projection : q1 > q2
(b) Time of flight : T1 > T2
(c) Horizontal range : R1 > R2
(d) Total energy : U1 > U2 .
Two particles are projected in air with speed vo at angles θ1 and θ2 (both acute) to the horizontal, respectively. If the height reached by the first particle is greater than that of the second, then tick the right choices
(a) Angle of projection : q1 > q2
(b) Time of flight : T1 > T2
(c) Horizontal range : R1 > R2
(d) Total energy : U1 > U2 .
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1 Answer
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This is a Multiple Choice Questions as classified in NCERT Exemplar
Answer- a,b,c
Explanation – H=
H1=Vo2sin2 1/2g , H2=Vo2sin2 2/2g
H1>H2
Vo2sin2 1/2g= Vo2sin2 2/2g
Sin2 1>sin2 2
Sin2 1 – sin2 2>0
(Sin 1 – sin 2)( Sin 1 + sin 2)>0
Sin 1>sin 2 or 1 >2
T=
T1= , T2=
T1> T2
R=
Sin 1>sin 2
Sin2 1> sin2 2
R1>R2
Total energy for the first particle
U1=K.E+P.E=1/2m1
U2= K.E+P.E= 1/2m2
Total energy for the second particle
So m1= m2 then U1=U2
So m1>m2 then U1>U2
So m1
2 then U1
Similar Questions for you
Please find the solution below:
after 10 kicks,
v? = 3tî v? = 24cos 60°î + 24sin 60°? = 12î + 12√3?
v? = v? – v? = (12 – 3t)î + 12√3?
It is minimum when 12 - 3t = 0 ⇒ t = 4sec
ω = θ² + 2θ
α = (ωdω)/dθ = (θ² + 2θ) (2θ + 2)
At θ = 1rad.
ω = 3rad/s and α = 12rad/s²
a? = αR = 12 m/s² a? = ω²R = 9 m/s² A? = √ (a? ² + a? ²) = 15 m/s²
a? = v? ²/4r
a_A? = (v? ²/r²) × r = v? ²/r
a_A = 3v? ²/4r
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