Motion in a Plane

To analyze the problem, consider two objects:First Object: Maximizing RangeTo achieve maximum range, the object must be projected at a $45^{\circ}$ angle.Using the formula for maximum height $H_1$ when launched at an angle $\theta_1=45^{\circ}$ :$H_1=\;\frac{u_1^2 \sin^2 \theta_1}{2g}=\;\frac{u_1^2 \times 1}{2g \times 2}=\;\frac{u_1^2}{4g}$Second Object: Maximizing HeightTo achieve maximum height, the object should be projected vertically at a $90^{\circ}$ angle.The formula for the maximum height $H_2$ with a launch angle $\theta_2=90^{\circ}$ :$H_2=\;\frac{u_2^2 \sin^2 90^{\circ}}{2g}=\;\frac{u_2^2}{2g}$Equating HeightsSince both objects reach the same maximum height:$H_1=H_2$From equations (i) and (ii):$\;\frac{u_1^2}{4g}=\;\frac{u_2^2}{2g}$Simplifying, we find:$u_1^2=2 u_2^2$Taking the square root:$u_1=\sqrt{2} u_2$Thus, the ratio of their initial velocities is:$\;\frac{u_1}{u_2}=\;\frac{\sqrt{2}}{1}$Therefore, the ratio of initial velocities is $\sqrt{2}:1$.