Question

Two particles are projected from ground
with same initial velocities at angles
60 and 30° (with horizontal). Let R, and
R, be their horizontal ranges, H, and
H, their maximum heights and T; and
T) are the time of fights. Then
(a)
Man​

Answer 1

Two particles are projected from the ground with same initial velocities at angles 60° and 30° with horizontal. Let R₁ and R₂ be their horizontal ranges, H₁ and H₂ be their maximum heights and T₁ and T₂ are their time of flights.

we have to find the relation between them.

solution : let initial velocity = u

here, R₁ = u²sin2(60°)/g = u²sin120°/g

= u²(√3/2)/g .....(1)

R₂ = u²sin2(30°)/g = u²(√3/2)/g .....(2)

here it is clear that R₁ = R₂ = R (let)

again, H₁ = u²sin²60°/2g = u²(3/4)/2g .... (3)

H₂ = u²sin²30°/2g = u²(1/4)/2g ........(4)

multiplying equations (3) and (4) we get,

H₁ × H₂ = (u²)²(3/4)(1/4)/(2g)²

⇒16 H₁ × H₂ = [u²(3/√2)/g]²

⇒16H₁H₂ = R²

Therefore, R₁ = R₂ = 4√(H₁H₂)

again, time of flight, T₁ = 2usin60°/g = √3u/g

time of flight, T₂ = 2usin30°/g = u/g

T₁T₂ = u²(√3)/g² = 2[u²(√3/2)/g]/g = 2R/g

Therefore, R = g/2 (T₁T₂)