Question

2. A ball is thrown at an angle theta with the horizontal.
Its horizontal range is equal to its maximum
height. This is possible only when the value of
tan theta is​

Answer 1

Answer:

please mark the answer as brainlest.

Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

Ball is thrown at an angle ${ \theta}$

Range of projectile = Maximum Height of projectile.

$\huge{\bold{\underline{Explanation:-}}}$

As range equals the Maximum height then ,

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

and,

$\large{R = \frac{2{u}^2 {\sin} \theta \cos \theta}{g}}$

$\large{ R = H_{max}}$

$\large{ \implies \frac{2{u}^2 {\sin} \theta \cos \theta}{g} = \frac{{u}^2 {\sin}^2 \theta}{2g}}$

On both sides u² , one sinθ and g will get cancel

It becomes,

$\large{ \implies 2 \cos \theta = \frac{ \sin \theta}{2}}$

$\large{ \implies \tan \theta = 4}$

So,the value of tanθ will be 4 when the projectile Has Maximum height = Range.