A ball is thrown at different angles with the same speed u and from the same points and it has same range in both the cases. If y1 and y2 be the heights attained in the two cases, then y1 + y2 =

The correct option is u²/2g.

Step-by-step Explanation:

For a projectile launched with an initial velocity u at an angle $\theta$ to the horizontal, the horizontal range R is given by the formula:

$R=\frac{u^{2}sin\left(2\theta \right)}{g}$

It is a well-known property of projectile motion that for a given initial speed u, the range is the same for complementary launch angles. Thus, if the range is the same in both cases, the two projection angles must be:

${\theta }_1=\theta$
and
${\theta }_2=90°-\theta$

The maximum height attained by a projectile is given by the formula:

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

Let $y_1$ and $y_2$ be the heights attained in the two cases. Substituting the angles, we get:

$y_1=\frac{u^2{sin}^2\theta }{2g}$

and

$y_2=\frac{u^2{sin}^2(90°-\theta )}{2g}$

Using the trigonometric identity $sin(90°-\theta )=cos\theta$, we can rewrite $y_2$ as:

$y_2=\frac{u^2{cos}^2\theta }{2g}$

Now, we find the sum of the two heights:

$y_1+y_2=\frac{u^2{sin}^2\theta }{2g}+\frac{u^2{cos}^2\theta }{2g}$

Factoring out the common terms:

$y_1+y_2=\frac{u^2}{2g}({sin}^2\theta +{cos}^2\theta )$

Using the fundamental trigonometric identity ${sin}^2\theta +{cos}^2\theta =1$, the equation simplifies to:

$y_1+y_2=\frac{u^2}{2g}$