Question

A ball is thrown at different angles with the same speed $u$ and from the same points and it has the same range in both cases. If $y_1$ and $y_2$ be the heights attained in the two cases. Then find the value of $y_1+y_2$.

• A. $\frac{u^2}{3g}$
• B. $\frac{u^2}{2g}$
• C. $\frac{u^2}{4g}$
• D. $\frac{u^3}{2g}$

Answer 1

The correct option is B $\frac{u^2}{2g}$

$\text{Step 1: Drawing projectile}$ $\text{[Ref. Fig.]}$

Maximum height for projectile is given by $H=\frac{u^2{\sin }^2\theta }{2g}$

$\text{Step 2: Range of projectile}$

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

Range at projection angle $\theta :$

$R_1=\frac{u^2\sin 2\theta }{g}$

Range at projection angle$90-\theta :$

$R_2=\frac{u^2\sin 2(90-\theta )}{g}$ $=\frac{u^2\sin 2\theta }{g}$

$\therefore R_1=R_2=R$

Therefore, Range is equal at $\theta$ and $90-\theta$

$\text{Step 3: Solving equation}$

For Projectile $1$, Maximum Height $y_1$

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

For Projectile $2$, Maximum Height $y_2$

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

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

Hence $y_1+y_2=\frac{u^2}{2g}$, Option B is correct.