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(A) The initial velocity is given by $\vec{u} = a \hat{i} + b \hat{j}$.Horizontal component of velocity,$u_x = a$.Vertical component of velocity,$u_y

Vedclass · Accepted Answer

$b = 2a$

Vedclass · Answer

(A) The initial velocity is given by $\vec{u} = a \hat{i} + b \hat{j}$.Horizontal component of velocity,$u_x = a$.Vertical component of velocity,$u_y = b$.The maximum height reached by the projectile is $H = \frac{u_y^2}{2g} = \frac{b^2}{2g}$.The horizontal range of the projectile is $R = \frac{2 u_x u_y}{g} = \frac{2ab}{g}$.According to the problem,the range is twice the maximum height: $R = 2H$.Substituting the expressions: $\frac{2ab}{g} = 2 \left( \frac{b^2}{2g} ight)$.Simplifying the equation: $\frac{2ab}{g} = \frac{b^2}{g}$.Dividing both sides by $b/g$ (assuming $b 
eq 0$): $2a = b$,or $b = 2a$.

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

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