Question

Two particles are projected with same speed in different directions in the same vertical plane. Both particles have same range. If their time of flights are 10 second and 25 seconds, then range is (g=10 m/s2)
(1) 1250 m
(2) 625 m
(3) 2500 m
(4) 312.5 m

Answer 1

Let two particles are projected with same speed u in different directions in the same vertical plane.
Let angle of projections of two particle with same vertical plane are A and B.

range of 1st particle = u²sin2A/g
range of 2nd particle = u²sin2B/g

A/C to question,
u²sin2A/g = u²sin2B/g
=> sin2A = sin2B
=> sin2A = sin(180° - 2B)
=> 2A = 180° - 2B
=> 2A + 2B = 180°
=> A + B = 90°

time of flight = 2usinA/g
10 = 2usinA/g
usinA = 50 .......(i)
time of flight = 2usinB/g
25 = 2usinB/g
125 = usinB
usinB = usin(90-A) = 125
ucosA = 125.........(ii)

from equations (i) and (ii),
usinA/ucosA = 50/125 = 2/5
tanA = 2/5
so, sinA = 2/√29
cosA = 5/√29
sin2A = 2cosA.sinA = 2 × 2/√29 × 5/√29
sin2A = 20/29

usinA = 50
u × 2/√29 = 50
u = 25√29

Range of 1st particle = range of 2nd particle = u²sin2A/g
= (25√29)² × 20/29/10
= 625 × 29 × 20/29/10
= 1250m

hence option (A) is correct

Answer 2

Let the both particles  are projected   with velocity of  v  and  and  angle  of α and β respectively,

For, First body,

$R_1=\frac{u^{2}\sin2\alpha}{g}\\ \\ T_1=\frac{2u\sin\alpha}{g}.........(1)$

For  second body,

$R_2=\frac{u^{2}\sin2\beta}{g}\\ \\ T_2=\frac{2u\sin\beta}{g}.........(2)$

According  to question,

$R_1=R_2\\ \\u^{2}\sin2\alpha}{g}=u^{2}\sin2\beta}{g}\\ \\\sin2\alpha=\sin2\beta\\ \\\sin2\alpha=\sin(180-2\beta)\\ \\2\alpha=180-2\beta\\ \\ \alpha=90-\beta$

T₁/T₂ = sinα/sinβ

10/25 = sinα/sin(90-α)

2/5 = sinα/cosα

tanα = 2/5

sinα = 2/√29

cosα = 5/√29

T₁ = 2usinα/g

$10=\frac{2u\frac{2}{\sqrt{29}}}{10}\\ \\u=\frac{25}{\sqrt{29}}\\ \\ R_1=R_2=\frac{u^{2}\sin2\alpha}{g}\\ \\=\frac{25\times25\times2\sin\alpha.\cos\alpha}{29\times10}=\frac{25\times25\times2\frac{2}{\sqrt{29}}.\frac{5}{\sqrt{29}}}{29\times10}=1250\;m$