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MCQEasyJEE 2025Projectile Motion

JEE Physics 2025 Question with Solution

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

  • A
    Graph of maximum height $$h_m$$ versus angle $$\phi$$ from $$0$$ to $$90^\circ$$, starting at origin and rising upward convexly, with a dashed vertical line near $$90^\circ$$.
  • B
    Graph of maximum height $$h_m$$ versus angle $$\phi$$ from $$0$$ to $$90^\circ$$, starting at zero, rising to a peak at intermediate $$\phi$$, then falling back to zero.
  • C
    Graph of maximum height $$h_m$$ versus angle $$\phi$$ from $$0$$ to $$90^\circ$$, starting at maximum value when $$\phi=0$$ and decreasing to zero at $$90^\circ$$ with downward curvature.
  • D
    Graph of maximum height $$h_m$$ versus angle $$\phi$$ from $$0$$ to $$90^\circ$$, starting at maximum value and decreasing steeply at first, then flattening toward zero near $$90^\circ$$.

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The angle of projection is measured from the vertical as ϕ\phi, and the maximum height is hmh_m.

Find: The correct graph of hmh_m as a function of ϕ\phi.

Projectile launched at an angle with the vertical, showing the velocity direction and parabolic trajectory used to identify the vertical component of initial velocity.

For projectile motion, maximum height depends on the initial vertical component of velocity.

Since the angle is measured from the vertical, the vertical component is ucos⁡ϕu\cos\phi.

Using the maximum height formula,

Hmax⁡=u2cos⁡2ϕ2gH_{\max} = \frac{u^2 \cos^2 \phi}{2g}

Therefore,

hm=u2cos⁡2ϕ2gh_m = \frac{u^2 \cos^2 \phi}{2g}

As ϕ\phi increases from 00 to 90∘90^\circ, cos⁡2ϕ\cos^2\phi decreases from 11 to 00. Hence hmh_m decreases continuously from its maximum value to zero.

Therefore, the correct option is C.

Angle Conversion Method

Given: The projectile is projected at angle ϕ\phi with the vertical.

Find: The relation between hmh_m and ϕ\phi.

The standard formula for maximum height when the projection angle with the horizontal is θ\theta is

hm=(v0sin⁡θ)22gh_m = \frac{(v_0\sin\theta)^2}{2g}

Here, the angle is given with the vertical, so

θ=90∘−ϕ\theta = 90^\circ - \phi

Substituting this into the formula,

hm=(v0sin⁡(90∘−ϕ))22gh_m = \frac{\left(v_0\sin(90^\circ-\phi)\right)^2}{2g}

Using

sin⁡(90∘−ϕ)=cos⁡ϕ\sin(90^\circ-\phi) = \cos\phi

we get

hm=(v0cos⁡ϕ)22g=v02cos⁡2ϕ2gh_m = \frac{(v_0\cos\phi)^2}{2g} = \frac{v_0^2\cos^2\phi}{2g}

Thus hmh_m decreases from maximum at ϕ=0\phi=0 to zero at ϕ=90∘\phi=90^\circ, which matches Option C.

Common mistakes

  • Using usin⁡ϕu\sin\phi as the vertical component is incorrect because ϕ\phi is measured from the vertical, not the horizontal. Use ucos⁡ϕu\cos\phi instead.

  • Choosing a graph that rises first and then falls is incorrect because cos⁡2ϕ\cos^2\phi decreases monotonically on 0≤ϕ≤90∘0\le\phi\le90^\circ. The height does not increase with ϕ\phi in this interval.

  • Confusing the standard angle θ\theta with the given angle ϕ\phi leads to a wrong substitution. First write θ=90∘−ϕ\theta=90^\circ-\phi before applying the projectile formula.

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