Question 281: Work, Energy and Power & Circular Motion

Question

A projectile is thrown with an initial velocity $$(a \hat{i}+b \hat{j}) \mathrm{m} / \mathrm{s}$$, 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

Accepted Answer

$$\mathrm{u}_{\mathrm{x}}=\mathrm{a}=$$ Horizontal component of the velocity
$$\mathrm{u}_{\mathrm{y}}=\mathrm{b}=$$ Vertical component of the velocity
Maximum height $$\mathrm{H}=\frac{\mathrm{u}_{\mathrm{y}}^2}{2 \mathrm{~g}}=\frac{\mathrm{b}^2}{2 \mathrm{~g}}$$
$$\begin{aligned}
	ext { Range } \mathrm{R} & =\frac{2 \mathrm{u}_{\mathrm{y}} \mathrm{u}_{\mathrm{x}}}{\mathrm{g}}=\frac{2 \mathrm{ba}}{\mathrm{g}} \
\mathrm{R} & =2 \mathrm{H} \quad 	herefore \frac{2 \mathrm{ba}}{\mathrm{g}}=\frac{2 \mathrm{~b}^2}{2 \mathrm{~g}} \
	herefore \quad \mathrm{b} & =2 \mathrm{a}
\end{aligned}$$

Answer

$$\mathrm{b}=2 \mathrm{a}$$

Answer

$$\mathrm{b}=4 \mathrm{a}$$

Answer

$$\mathrm{b=\frac{a}{2}}$$

Answer

$$\mathrm{b}=\mathrm{a}$$

Work, Energy and Power & Circular Motion

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

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