Question

Two balls are projected at an angle ϴ and (90º – ϴ) to the
horizontal with the same speed. The ratio of their maximum
vertical heights is
(a) 1 : 1 (b) tanϴ : 1
(c) 1 : tanϴ (d) tan2ϴ : 1

Answer 1

Two balls are projected at an angle θ and (90° — θ) to the horizontal with the same speed.

Now,

The maximum vertical height is given by

H = (u² sin²θ)/2g

The first ball is projected at an angle θ.

So,

H = (u²sin²θ)/2g

Similarly, the second ball is protected at an angle (90° - θ)

H = [u² sin²(90° - θ)]/2g

We have to find the ratios of maximum vertical height.

So, divide them (ratio of the first ball by the second ball)

[(u²sin²θ)/2g] / [{u² sin²(90° - θ)}/2g]

[(u²sin²θ)/2g] / [(u²cos²θ)/2g]

sin²θ/cos²θ

tan²θ/1

Therefore, the ratio of their maximum vertical heights is tan²θ:1.

Option d) tan²θ:1

Answer 2

GIVEN:

• Two balls projected at angles θ & (90° - θ)

TO FIND:

• Ratio of maximum height

SOLUTION:

Using the formula

H = u²sin²θ/2g

Given the first ball projected in angle θ

H = u²sin²θ/2g

Assuming H₁

_________________________________

And second ball projected in angel 90° - θ

H = u²sin²(90° - θ)/2g

Substituting sin(90° - θ) = cosθ

H = u²cos²θ/2g

Assuming as H₂

Now, we need to find H1 : H2

→ H₁ : H₂ = (u²sin²θ/2g)/u²cos²θ/2g

→ H₁ : H₂ = sin²θ/cos²θ

→ H₁ : H₂ = tan²θ : 1 ( Option D )

Hence, option D is your answer