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:
- Upward motion (against gravity): The body slows down due to gravity until its velocity becomes zero.
- 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:
- $v = u + at$
- $s = ut + \frac{1}{2}at^2$
- $v^2 = u^2 + 2as$
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:
- Rise to a maximum height where its velocity becomes zero.
- 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
| Quantity | Formula |
|---|---|
| Time to reach max height $t_1$ | $\frac{u}{g}$ |
| Max height above throw point | $\frac{u^2}{2g}$ |
| Total height from ground H | $h + \frac{u^2}{2g}$ |
| Time to fall from max height | $\sqrt{\frac{2(h + \frac{u^2}{2g})}{g}}$ |
| Total time of flight T | $\frac{u}{g} + \sqrt{\frac{2(h + \frac{u^2}{2g})}{g}}$ |
| Velocity before hitting ground | $\sqrt{2g(h + \frac{u^2}{2g})}$ |
🔄 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:
- Maximum height above the ground
- Time of flight
- 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:
$v = \sqrt{2g(h + h_1)} = \sqrt{2 \times 9.8 \times 65.41} ≈ \sqrt{1289.8} ≈ 35.9 \, \text{m/s}$
