Physics Q83 – AP-EAMCET 2025

Home Physics Motion in a Straight LineQuestion #317

If a ball released from a height H takes a time T to reach the ground, then the position of the ball from the ground at a time

\frac{T}{2}

Clue 1 of 3: Kinematics of Free Fall

When an object is released from rest and falls under gravity, its motion can be described by the equations of motion with constant acceleration due to gravity.

🧠 Full Solution Path

Step 1: Determine the total height in terms of total time✦ Active

The ball is released from height $H$ , meaning its initial velocity $u = 0$ . It takes time $T$ to reach the ground. Using the equation of motion for free fall, $s = u t + \frac{1}{2} g t^{2}$ .

H = (0) T + \frac{1}{2} g T^{2}

H = \frac{1}{2} g T^{2} (1)

💡 Teacher's Secret Hint

Remember that 'released' implies an initial velocity of zero. The acceleration due to gravity $g$ is constant and acts downwards.

Step 2: Calculate the distance fallen from the top at time

\frac{T}{2}

○ Expand

Let $h^{'}$ be the distance fallen from the top at time $t^{'} = \frac{T}{2}$ . Using the same equation of motion:

h^{'} = (0) t^{'} + \frac{1}{2} g (t^{'})^{2}

h^{'} = \frac{1}{2} g {(\frac{T}{2})}^{2}

h^{'} = \frac{1}{2} g \frac{T^{2}}{4}

h^{'} = \frac{1}{4} (\frac{1}{2} g T^{2}) (2)

💡 Teacher's Secret Hint

Be careful with squaring the time. $(\frac{T}{2})^{2}$ becomes $\frac{T^{2}}{4}$ , not $\frac{T^{2}}{2}$ .

Step 3: Express the distance fallen in terms of H○ Expand

Substitute equation (1) into equation (2):

h^{'} = \frac{1}{4} H

This $h^{'}$ is the distance the ball has fallen *from the top* at time $\frac{T}{2}$ .

💡 Teacher's Secret Hint

This step simplifies the expression by relating the distance fallen to the total height, making the final calculation easier.

Step 4: Determine the position of the ball from the ground○ Expand

The question asks for the position of the ball *from the ground*. Let this be $y$ . This is the total height minus the distance fallen from the top.

y = H - h^{'}

y = H - \frac{H}{4}

y = \frac{4 H - H}{4}

y = \frac{3 H}{4}

Therefore, the position of the ball from the ground at time $\frac{T}{2}$ is $\frac{3 H}{4}$ .

💡 Teacher's Secret Hint

Always re-read the question to ensure you are calculating what is asked. Sometimes, the distance fallen from the top is asked, and other times, the height from the ground.

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Physics Question 83 – AP-EAMCET 2025

Clue 1 of 3: Kinematics of Free Fall

🧠 Full Solution Path