Kinematics

JEE Main level, single correct, medium

198 of 308

Question 198 of 308

Two identical bodies, projected with the same speed at two different angles cover the same horizontal range RR. If the time of flight of these bodies are 55 s and 1010 s, respectively, then the value of RR is ________ m. (Take g=10g=10 m/s2^2)

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Answer: (A) 250250

For two projectiles to have the same horizontal range RR with the same initial speed uu, their angles of projection must be complementary, i.e., θ\theta and 90∘−θ90^\circ - \theta.

The times of flight for these two angles are given by:
t1=2usin⁡θgt_1 = \dfrac{2u \sin \theta}{g}
t2=2ucos⁡θgt_2 = \dfrac{2u \cos \theta}{g}

Multiplying t1t_1 and t2t_2, we get:
t1t2=(2usin⁡θg)(2ucos⁡θg)=2g(u2(2sin⁡θcos⁡θ)g)t_1 t_2 = \left(\dfrac{2u \sin \theta}{g}\right) \left(\dfrac{2u \cos \theta}{g}\right) = \dfrac{2}{g} \left(\dfrac{u^2 (2 \sin \theta \cos \theta)}{g}\right)

Since the horizontal range is R=u2sin⁡2θgR = \dfrac{u^2 \sin 2\theta}{g}, we can write:
t1t2=2Rgt_1 t_2 = \dfrac{2R}{g}

Given t1=5t_1 = 5 s, t2=10t_2 = 10 s, and g=10g = 10 m/s2^2, substituting these values:
5×10=2R105 \times 10 = \dfrac{2R}{10}

50=2R1050 = \dfrac{2R}{10}

2R=500⇒R=2502R = 500 \Rightarrow R = 250 m

Answer: 250250

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