Physics Q27 – JEE-MAIN 2025

Home Physics Motion in a PlaneQuestion #308

A particle is projected with velocity

u

so that its horizontal range is three times the maximum height attained by it. The horizontal range of the projectile is given as

\frac{n u^{2}}{25 g}

, where value of

n

is: (Given, '

g

' is the acceleration due to gravity.)

Clue 1 of 3: Projectile Motion Basics

Recall the standard formulas for the horizontal range and maximum height of a projectile.

🧠 Full Solution Path

Step 1: Express Range and Height in terms of initial velocity and angle✦ Active

The horizontal range ( $R$ ) and maximum height ( $H$ ) of a projectile launched with initial velocity $u$ at an angle $θ$ with the horizontal are given by:

R = \frac{u^{2} \sin (2 θ)}{g}

H = \frac{u^{2} \sin^{2} θ}{2 g}

Step 2: Use the given condition to find the angle of projection○ Expand

Given that the horizontal range is three times the maximum height, we have $R = 3 H$ . Substitute the formulas for $R$ and $H$ :

\frac{u^{2} \sin (2 θ)}{g} = 3 (\frac{u^{2} \sin^{2} θ}{2 g})

Simplify the equation using $\sin (2 θ) = 2 \sin θ \cos θ$ :

\frac{u^{2} (2 \sin θ \cos θ)}{g} = \frac{3 u^{2} \sin^{2} θ}{2 g}

Cancelling common terms ( $u^{2} / g$ ) and assuming $\sin θ \neq 0$ :

2 \cos θ = \frac{3}{2} \sin θ ⟹ \tan θ = \frac{4}{3}

💡 Teacher's Secret Hint

Remember to use the double angle identity for sine, $\sin (2 θ) = 2 \sin θ \cos θ$ .

Step 3: Calculate the range and determine the value of n○ Expand

From $\tan θ = \frac{4}{3}$ , we can construct a right triangle to find $\sin θ = \frac{4}{5}$ and $\cos θ = \frac{3}{5}$ . Now, calculate $\sin (2 θ)$ :

\sin (2 θ) = 2 \sin θ \cos θ = 2 (\frac{4}{5}) (\frac{3}{5}) = \frac{24}{25}

Substitute this value back into the range formula:

R = \frac{u^{2} \sin (2 θ)}{g} = \frac{u^{2}}{g} (\frac{24}{25}) = \frac{24 u^{2}}{25 g}

Comparing this with the given expression for range, $R = \frac{n u^{2}}{25 g}$ , we find:

n = 24

💡 Teacher's Secret Hint

Ensure you correctly derive $\sin θ$ and $\cos θ$ from $\tan θ$ .

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Physics Question 27 – JEE-MAIN 2025

Clue 1 of 3: Projectile Motion Basics

🧠 Full Solution Path