Question

what is ratio of maximum height attained by object when it thrown with complementary angles​

Answer 1

Given that,

The maximum height attained by objects when it thrown with complementary angles.

Let the angle of one object is θ and other angle is (90-θ ).

We know that,

The maximum height attained by the projectile is

$H=\dfrac{u^2\sin\theta}{2g}$

We need to calculate the ratio of maximum height attained by object

Using formula of height

$\dfrac{H_{1}}{H_{2}}=\dfrac{\dfrac{u^2\sin\theta}{2g}}{\dfrac{u^2\sin(90-\theta)}{2g}}$

$\dfrac{H_{1}}{H_{2}}=\dfrac{\sin\theta}{\sin(90-\theta)}$

$\dfrac{H_{1}}{H_{2}}=\dfrac{\sin\theta}{\cos\theta}$

$\dfrac{H_{1}}{H_{2}}=\tan\theta$

Hence, The ratio of maximum height attained by object is tan θ.