Two spheres are projected at angles 30^{circ} | vedclass

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(B) The formula for the maximum height $H$ reached by a projectile is given by $H = \frac{u^2 \sin^2 	heta}{2g}$, where $u$ is the initial velocity a

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$\sqrt{2}:1$

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(B) The formula for the maximum height $H$ reached by a projectile is given by $H = \frac{u^2 \sin^2 	heta}{2g}$, where $u$ is the initial velocity and $	heta$ is the angle of projection.Given that the maximum heights are equal for both spheres, we have $H_1 = H_2$.Therefore, $\frac{u_1^2 \sin^2 	heta_1}{2g} = \frac{u_2^2 \sin^2 	heta_2}{2g}$.This simplifies to $u_1^2 \sin^2 	heta_1 = u_2^2 \sin^2 	heta_2$.Substituting the given values $	heta_1 = 30^{\circ}$ and $	heta_2 = 45^{\circ}$:$u_1^2 \sin^2(30^{\circ}) = u_2^2 \sin^2(45^{\circ})$.$u_1^2 (0.5)^2 = u_2^2 (1/\sqrt{2})^2$.$u_1^2 (1/4) = u_2^2 (1/2)$.$u_1^2 / u_2^2 = (1/2) / (1/4) = 4/2 = 2$.Taking the square root on both sides, we get $u_1 / u_2 = \sqrt{2} / 1$.Thus, the ratio of their initial velocities is $\sqrt{2}:1$.

$(A)$ Range	$(i)$ $\frac{P}{Q}$
$(B)$ Maximum height	$(ii)$ $P$
$(C)$ Time of flight	$(iii)$ $\frac{P^2}{4 Q}$
$(D)$ Tangent of projection	$(iv)$ $\left(\sqrt{\frac{2}{g Q}}\right) P$

Two spheres are projected at angles $30^{\circ}$ and $45^{\circ}$ with the horizontal. The maximum height reached by both is same. The ratio of their initial velocities is, $(\sin 45^{\circ} = \frac{1}{\sqrt{2}}, \sin 30^{\circ} = 0.5)$

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