A ball of mass 1 ; kg is thrown vertically up | vedclass

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(B) For the first ball thrown vertically upwards,the time of flight is $T_1 = \frac{2u_1}{g} = 3 \; s$. Thus,$u_1 = \frac{3g}{2}$. The maximum height

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$1:1$

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(B) For the first ball thrown vertically upwards,the time of flight is $T_1 = \frac{2u_1}{g} = 3 \; s$. Thus,$u_1 = \frac{3g}{2}$. The maximum height is $H_1 = \frac{u_1^2}{2g} = \frac{(3g/2)^2}{2g} = \frac{9g}{8}$.For the second ball thrown at an angle $	heta = 60^{\circ}$ with the vertical,the angle with the horizontal is $\alpha = 90^{\circ} - 60^{\circ} = 30^{\circ}$. The time of flight is $T_2 = \frac{2u_2 \sin \alpha}{g} = 3 \; s$. Thus,$u_2 \sin \alpha = \frac{3g}{2}$.The maximum height is $H_2 = \frac{(u_2 \sin \alpha)^2}{2g} = \frac{(3g/2)^2}{2g} = \frac{9g}{8}$.Comparing the two heights,$\frac{H_1}{H_2} = \frac{9g/8}{9g/8} = 1:1$.

$A$ ball of mass $1 \; kg$ is thrown vertically upwards and returns to the ground after $3 \; s$. Another ball,thrown at $60^{\circ}$ with the vertical,also stays in the air for the same time before it touches the ground. The ratio of the two maximum heights reached is:

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