Question 11: Motion in 2D & Vectors

Question

Two balls with same mass and initial velocity, are projected at different angles in such a way that maximum height reached by first ball is 8 times higher than that of the second ball. $T_1$ and $T_2$ are the total flying times of first and second ball, respectively, then the ratio of $T_1$ and $T_2$ is

Accepted Answer

Given that the maximum height reached by the first ball is 8 times that of the second ball:
$ \left(H_{\max}ight)_1 = 8 	imes \left(H_{\max}ight)_2 $
We can relate this to the initial velocities and angles using the formula for maximum height:
$ \frac{u^2 \sin^2 	heta_1}{2g} = 8 	imes \frac{u^2 \sin^2 	heta_2}{2g} $
Simplifying, we find:
$ \sin 	heta_1 = 2 \sqrt{2} \sin 	heta_2 $
Now, the ratio of the total flying times $T_1$ and $T_2$ is given by:
$ \frac{T_1}{T_2} = \frac{2u \sin 	heta_1 / g}{2u \sin 	heta_2 / g} = \frac{\sin 	heta_1}{\sin 	heta_2} = 2 \sqrt{2} $
Thus, the ratio of $T_1$ to $T_2$ is $2 \sqrt{2} : 1$.

Answer

2 : 1

Answer

$ \sqrt{2} : 1 $

Answer

2$ \sqrt{2} : 1 $

Answer

4 : 1

Motion in 2D & Vectors

Two balls with same mass and initial velocity, are projected at different angles in such a way that maximum height reached by first ball is 8 times higher than that of the second ball. T_1 and T_2 are the total flying times of first and second ball, respectively, then the ratio of T_1 and T_2 is

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