Question

Prove that for two angles of projection and (90- ), with the same initial velocity ‘u’, (i) range is same, (ii) heights are in the ratio tan 2 :1

Answer 1

Solution:

(i)

• R = (u²sin2Ø)/g

For angle Ø,

R1 = (u²sin2Ø)/g ...(1)

For angle 90 - Ø,

R2 = [u²sin2(90 - Ø)]/g

=> R2 = [u²sin(180 - 2Ø)]/g

=> R2 = (u²sin2Ø)/g ...(2)

For (1) and (2),

Range is same for Ø and (90 - Ø)

Note:

sin (180 - 2Ø) = sin2Ø

Reason Ø is an acute angle therefore Ø < 90⁰ and 2Ø < 180⁰ since sin is positive in first two quadrant sin 2Ø will be positive.

(ii)

• H = (u²sin²Ø)/2g

For angle Ø,

H1 = (u²sin²Ø)/2g...(1)

For angle 90 - Ø,

H2 = [u²sin²(90 - Ø)]/2g...(2)

On dividing one and 2 we get

H1/H2 = sin²Ø/sin²(90 - Ø)

=> H1/H2 = sin²Ø/cos²Ø

=> H1/H2 = tan²Ø

Therefore, the ratio of heights is tan²Ø : 1.