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← JEE Advanced 2024 · Paper 2
PhysicsInteger+4Question 26 of 51

JEE Advanced 2024 · Paper 2, Physics question 9

A ball is thrown from the location (x0,y0)=(0,0)(x_0, y_0) = (0, 0) of a horizontal playground with an initial speed v0v_0 at an angle θ0\theta_0 from the +x+x-direction. The ball is to be hit by a stone, which is thrown at the same time from the location (x1,y1)=(L,0)(x_1, y_1) = (L, 0). The stone is thrown at an angle (180−θ1)(180 - \theta_1) from the +x+x-direction with a suitable initial speed. For a fixed v0v_0, when (θ0,θ1)=(45∘,45∘)(\theta_0, \theta_1) = (45^{\circ}, 45^{\circ}), the stone hits the ball after time T1T_1, and when (θ0,θ1)=(60∘,30∘)(\theta_0, \theta_1) = (60^{\circ}, 30^{\circ}), it hits the ball after time T2T_2. In such a case, (T1/T2)2(T_1/T_2)^2 is _____.

Answer and worked solution

Official answer: 2

Both objects have the same acceleration gg, so their relative velocity is constant. The stone hits the ball only if their vertical velocity components are equal:
v0sin⁡θ0=usin⁡θ1.v_0\sin\theta_0=u\sin\theta_1.
The horizontal gap LL then closes at speed v0cos⁡θ0+ucos⁡θ1v_0\cos\theta_0+u\cos\theta_1.

(45∘,45∘)(45^\circ,45^\circ). u=v0u=v_0, so T1=L2 v0T_1=\dfrac{L}{\sqrt2\,v_0}.

(60∘,30∘)(60^\circ,30^\circ). u=v0sin⁡60∘sin⁡30∘=3 v0u=v_0\dfrac{\sin60^\circ}{\sin30^\circ}=\sqrt3\,v_0, so
T2=Lv02+3v0⋅32=L2v0.T_2=\frac{L}{\frac{v_0}{2}+\sqrt3v_0\cdot\frac{\sqrt3}{2}}=\frac{L}{2v_0}.

(T1T2)2=(22)2=2.\left(\frac{T_1}{T_2}\right)^2=\left(\frac{2}{\sqrt2}\right)^2=2.