Physics Q29 – JEE-MAIN 2026

Home Physics Motion in a Straight LineQuestion #322

If x and y coordinates of a projectile as a function of time (

t

) are given as

24 t

and

43.6 t - 4.9 t^{2}

, respectively, then the angle (in degrees) made by the projectile with horizontal when

t = 2

s is _______.

Clue 1 of 3: Projectile Motion Velocity

The angle a projectile makes with the horizontal at any instant is determined by the ratio of its vertical and horizontal velocity components at that instant.

🧠 Full Solution Path

Step 1: Calculate Velocity Components✦ Active

Given the position coordinates as functions of time:

x (t) = 24 t

y (t) = 43.6 t - 4.9 t^{2}

Differentiate $x (t)$ and $y (t)$ with respect to time to find the velocity components $v_{x} (t)$ and $v_{y} (t)$ :

v_{x} (t) = \frac{d x}{d t} = \frac{d}{d t} (24 t) = 24 m/s

v_{y} (t) = \frac{d y}{d t} = \frac{d}{d t} (43.6 t - 4.9 t^{2}) = 43.6 - 2 \times 4.9 t = 43.6 - 9.8 t m/s

Step 2: Evaluate Velocities at

t = 2

s○ Expand

Substitute $t = 2$ s into the velocity component equations:

v_{x} = 24 m/s

v_{y} = 43.6 - 9.8 (2) = 43.6 - 19.6 = 24 m/s

Step 3: Determine the Angle with Horizontal○ Expand

The angle $θ$ made by the projectile with the horizontal is given by $\tan θ = \frac{v_{y}}{v_{x}}$ :

\tan θ = \frac{24}{24} = 1

Therefore, the angle is:

θ = \arctan (1) = 45^{\circ}

💡 Teacher's Secret Hint

Remember that $\arctan (1)$ corresponds to $45^{\circ}$ or $\frac{π}{4}$ radians.

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Physics Question 29 – JEE-MAIN 2026

Clue 1 of 3: Projectile Motion Velocity

🧠 Full Solution Path