Physics Q47 – JEE-MAIN 2026

Home Physics Motion in a Straight LineQuestion #323

From

18 m

height above the ground a ball is dropped from rest. The height above the ground at which the magnitude of velocity equal to the magnitude of acceleration (in the same set of units) due to gravity is _______ m. (Take

g = 10 {m/s}^{2}

and neglect the air resistance)

Enter Calculated Numerical Value:

Clue 1 of 3: Kinematics of Free Fall

Understand the motion of an object dropped from rest under constant gravitational acceleration.

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Step 1: Identify Given Information and Condition✦ Active

The ball is dropped from rest, so initial velocity $u = 0$ . The initial height is $H = 18 m$ . The acceleration due to gravity is $g = 10 {m/s}^{2}$ . We need to find the height above the ground where the magnitude of velocity equals the magnitude of acceleration, which means $v = g = 10 m/s$ .

Step 2: Calculate Distance Fallen○ Expand

Using the kinematic equation $v^{2} = u^{2} + 2 g s$ , where $s$ is the distance fallen:

v^{2} = 0^{2} + 2 g s

Substitute $v = g$ into the equation:

g^{2} = 2 g s

Since $g \neq 0$ , we can divide by $g$ to find the distance fallen $s$ :

s = \frac{g}{2}

Substitute $g = 10 {m/s}^{2}$ :

s = \frac{10}{2} = 5 m

💡 Teacher's Secret Hint

Remember that 's' represents the displacement from the starting point, not the height from the ground.

Step 3: Calculate Height Above Ground○ Expand

The height above the ground $h$ is the initial height $H$ minus the distance fallen $s$ :

h = H - s

Substitute the values $H = 18 m$ and $s = 5 m$ :

h = 18 - 5 = 13 m

💡 Teacher's Secret Hint

Always double-check what the question is asking for: distance fallen or height above the ground.

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Physics Question 47 – JEE-MAIN 2026

Clue 1 of 3: Kinematics of Free Fall

🧠 Full Solution Path