JEE Main 2025 Apr 8 Shift 2, Physics Q16: Motion In Two And Three Dimension

JEE Main2025Apr 8, Shift 2Physics

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:

A
$22 : 1$
B
$2 : 1$
C
$2 : 1$
D
$4 : 1$

Solution

Maximum height of projectile: $H = \frac{u ^{2} sin ^{2} θ}{2 g}$ .

Given $H_{1} = 8 H_{2}$ with the same $u$ : $sin^{2} θ_{1} = 8 sin^{2} θ_{2}$ , so $sin θ_{1} = 22 sin θ_{2}$ .

Time of flight: $T = \frac{2 u sin θ}{g}$ , so $\frac{T _{1}}{T _{2}} = \frac{sin θ _{1}}{sin θ _{2}} = 22$ .

Therefore $T_{1} : T_{2} = 22 : 1$ , matching option (1).

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JEE Main 2025 Apr 8 Shift 2, Physics Q16: Motion In Two And Three Dimension

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