Two particles are projected in the air with speed \(v_0\), at angles \(\theta_1\) and \(\theta_2\) to the horizontal, respectively. If the height reached by the first particle is greater than that of the second, then:

(a) the angle of the projection: \(\theta_1>\theta_2\) 
(b) the time of flight: \(T_1>T_2\)
(c) the horizontal range: \(R_1>R_2\)
(d) the total energy: \(U_1>U_2\)

Choose the correct option:
1. (a), (c), (d)
2. (a), (c)
3. (b), (c), (d)
4. (a), (b)

(4) Hint: The maximum height depends on the angle of projection.

Step 1: Find a relation between angles.z

We know that maximum height reached by a projectile,

H=u2sin2θ2gH1=v02sin2θ12g                  for first particleH2=v02sin2θ22g                  for second particle

According to the question, we know that

H1>H22g>v02sin2θ12g>v02sin2θ22gsin2θ1>sin2θ2
sin2θ1sin2θ2>0(sinθ1sinθ2)(sinθ1+sinθ2)>0
 Thus, either sinθ1+sinθ2>0sinθ1sinθ2>0sinθ1>sinθ2 or θ1>θ2 Time of fight, T=2usinθg=2v0sinθg Thus, T1=2v0sinθ1gT2=2v0sinθ2g

(Here. T1 = Time of flight of first particle and T2 = Time of flight of second particle).

Step 2: Find relation between time of flight and the range.

 As, sinθ1>sinθ2 Hence, T1>T2

We know that,

 Range, R=u2sin2θg=v02sin2θgA1= Range of first particle =u02sin2θ1gR2= Range of second particle =v02sin2θ2g

Given,

 sinθ1>sinθ2sin2θ1>sin2θ2R1R2=sin2θ1sin2θ2>1R1>R2

Step 3: Find a relation between the total energies.

Total energy for the first particle.

U1=KE+PE=12m1v02 (This value will be constant throughout the journey) U2=KE+PE=12m2v02 (Total energy for the second particle) 

Total energy for the particle

m1=m2 then U1=U2m1>m2 then U1>U2m1<m2, then U1<U2