Question

A projectile is thrown with an initial velocity $(a \hat{i}+b \hat{j}) \text{ m/s}$, where $\hat{i}$ and $\hat{j}$ are unit vectors along horizontal and vertical directions respectively. If the range of the projectile is twice the maximum height reached by it, then

• A. $b = 2a$
• B. $b = 4a$
• C. $b = \frac{a}{2}$
• D. $b = a$

Hint:

For any projectile trajectory, there is a handy shortcut relating range and maximum height: $\tan\theta = \frac{4H}{R}$, where $\theta$ is the angle of projection. Here, since $R = 2H$, the formula yields $\tan\theta = \frac{4H}{2H} = 2$. Since $\tan\theta = \frac{u_y}{u_x} = \frac{b}{a}$, we directly get $\frac{b}{a} = 2 \implies b = 2a$.

Answer 1

The Correct Option is A

Step 1: Identify the components.
The launch velocity is $a\hat{i} + b\hat{j}$, so the horizontal component is $u_x = a$ and the vertical component is $u_y = b$. We are told the range $R$ is twice the maximum height $H$.
Step 2: Write the height formula.
Maximum height depends only on the vertical component: $H = \dfrac{u_y^2}{2g} = \dfrac{b^2}{2g}$.
Step 3: Write the range formula.
Range uses both components: $R = \dfrac{2 u_x u_y}{g} = \dfrac{2ab}{g}$.
Step 4: Apply the condition $R = 2H$.
$\dfrac{2ab}{g} = 2 \cdot \dfrac{b^2}{2g} = \dfrac{b^2}{g}$.
Step 5: Simplify.
Multiply both sides by $g$: $2ab = b^2$. Since $b \neq 0$ for a real projectile, divide by $b$: $2a = b$.
Step 6: State the relation.
Hence $b = 2a$, option (1).
\[ \boxed{b = 2a} \]