Body Thrown Vertically from a Height: Equations of Motion | JEE/NEET/CBSE Class 11

ANAND CLASSES Study Material and Notes to learn the equations of motion for a body thrown vertically from a height with detailed explanations, derivations, and solved examples. Ideal for JEE, NEET, and CBSE Class 11 Physics preparation.

🧠 Concept Overview

When a body is thrown vertically upward from a certain height, it undergoes vertical motion under the influence of gravity (g = 9.8 m/s² downward). This motion can be divided into two parts:

1. Upward motion (against gravity): The body slows down due to gravity until its velocity becomes zero.
2. Downward motion (with gravity): The body accelerates downward until it hits the ground.

📚 Assumptions and Sign Convention

• Acceleration due to gravity: g=9.8 m/s²
• The initial position is the height from which the body is thrown.
• Let the initial velocity of the body be u (positive if thrown upward).
• Let the height from ground be h.
• Let the displacement from the point of projection be s.

🧾 Equations of Motion (Under Gravity)

We use the three equations of motion:

1. $v = u + at$
2. $s = ut + \frac{1}{2}at^2$
3. $v^2 = u^2 + 2as$

Where:

• $u$ = initial velocity
• $v$ = final velocity
• $a$ = acceleration (negative for upward motion)
• $t$ = time
• $s$ = displacement

🔍 Body Thrown Vertically Upward from a Height

Let’s say a body is thrown vertically upward from a height h with initial velocity u. It will:

1. Rise to a maximum height where its velocity becomes zero.
2. Then fall back down past the initial height and eventually hit the ground.

📌 Step-by-Step Analysis

🔼 1. Upward Motion (until max height)

At max height:

• Final velocity $v = 0$
• Acceleration $a = -g$
• Using the equation of motion : $$v = u – gt$$ $$0 = u – gt_1$$ $$ t_1 = \frac{u}{g}$$

📌 Time to reach maximum height: $$t_1 = \frac{u}{g}$$

📌 Maximum height above projection point:
Using the equation of motion $$v^2 = u^2 – 2g h_1$$

$$0 = u^2 – 2g h_1$$

$$h_1 = \frac{u^2}{2g}$$

🔽 2. Downward Motion (from top to ground)

Now the body starts falling from a height: $$H = h + h_1 = h + \frac{u^2}{2g}$$

It starts from rest at height H and falls under gravity.

📌 Time to fall from maximum height to ground (t₂):

Using Equations of Motion : $$s = \frac{1}{2}gt^2$$

$$H = \frac{1}{2}gt_2^2 $$

$$ t_2 = \sqrt{\frac{2H}{g}} $$

$$ t_2 = \sqrt{\frac{2(h + \frac{u^2}{2g})}{g}}$$

🧮 Total Time of Flight

$$T = t_1 + t_2 = \frac{u}{g} + \sqrt{\frac{2(h + \frac{u^2}{2g})}{g}}$$

🧨 Velocity Just Before Hitting Ground

Use $v^2 = u^2 + 2gH$, where $u = 0$ (because it’s falling from rest at maximum height): $$v = \sqrt{2g(h + \frac{u^2}{2g})}$$

🧠 Quick Summary Table

🔄 Special Cases

📌 Case 1: Thrown Downward

• Take $u$ as positive, motion is in direction of gravity.
• Use standard equations: $s = ut + \frac{1}{2}gt^2$, $\quad v = u + gt$, $\quad v^2 = u^2 + 2gs$

📌 Case 2: Dropped from Height (Free Fall)

• $u$ = 0
• Use: $s = \frac{1}{2}gt^2$, $\quad v = gt$, $\quad v^2 = 2gs$

📝 Example Question

🔢 Q: A ball is thrown vertically upward with speed 20 m/s from a building 45 m high. Find:

1. Maximum height above the ground
2. Time of flight
3. Speed before hitting the ground

✅ Solution:

Given:
$u = 20 \, \text{m/s}, \quad h = 45 \, \text{m}, \quad g = 9.8 \, \text{m/s}^2$

$h_1 = \frac{u^2}{2g} = \frac{400}{2 \times 9.8} ≈ 20.41 \, \text{m} $

🔸 Total height = $45 + 20.41 = 65.41 \, \text{m}$

$t_1 = \frac{u}{g} = \frac{20}{9.8} ≈ 2.04 \, \text{s} $

$t_2 = \sqrt{\frac{2 \times 65.41}{9.8}} ≈ \sqrt{13.35} ≈ 3.65 \, \text{s}$

🔸 Total time = $t_1 + t_2 = 2.04 + 3.65 ≈ 5.69 \, \text{s}$

Velocity before hitting ground:

$v = \sqrt{2g(h + h_1)} = \sqrt{2 \times 9.8 \times 65.41} ≈ \sqrt{1289.8} ≈ 35.9 \, \text{m/s}$