The angle of projection of a particle is measured from the vertical axis as φ and the maximum height reached by the particle is hm. Here hm as function of φ can be presented as (a) (b) (c) (d)

Answer: (d). Angle from the vertical is φ, so the vertical component of velocity is ucos φ. hm = (u²cos ²φ)/(2g) At φ = 0 (straight up) hm is maximum, u²/2g, and it falls smoothly to zero at φ = 90°. The curve starts at its peak with zero slope and drops to zero at 90°: graph (d).

The angle of projection of a particle is measured from the vertical… | JEE Main 2025 (3 Apr, Shift 1)

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Time of Flight, Maximum Height & Range

The angle of projection of a particle is measured from the vertical axis as $\phi$ and the maximum height reached by the particle is $h_m$ . Here $h_m$ as function of $\phi$ can be presented as

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Answer: (d)

Angle from the vertical is $\phi$ , so the vertical component of velocity is $u\cos\phi$ .
$h_m = \frac{u^2\cos^2\phi}{2g}$
At $\phi = 0$ (straight up) $h_m$ is maximum, $\frac{u^2}{2g}$ , and it falls smoothly to zero at $\phi = 90^\circ$ .
The curve starts at its peak with zero slope and drops to zero at $90^\circ$ : graph (d).

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