Question

15. Two balls are projected at an angle θ and (90° — θ) to the horizontal with the same speed. The ratio of their maximum vertical heights is *

Answer 1

Answer:

Explanation:

maximum vertical height of angle θ  

=> u²sin²θ/2g

max vertical height of angle 90 - θ

=> u²sin²(90-θ)/2g

=> u²cos²θ/2g

Ratio = (u²sin²θ/2g) / (u²cos²θ /2g)

=> sin²θ/cos²θ

=> tan²θ/1

=> tan²θ:1

                                                       

Answer 2

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

Therefore,

The ratio of their maximum vertical heights is tan²θ:1