a stone is projected vertically upward with a speed of 30 ms⁻¹ from the top of a tower of height 50 m. neglecting air resistance, determine g = 10 ms⁻²
a. the maximum height reached from the ground.
b. the time of flight
c. the total distance

question 2
a. sketch velocity - time graph to illustrate each of the following:
i) uniform motion;
ii) uniform acceleration with an initial velocity u = a m/s (a 1)
iii) a body moving with zero acceleration from point a to point b and then decelerate to point c
b. with an illustration how can we determine the
i) the acceleration
ii) deceleration
iii) total distance

question 3
a. a particle moving in a straight line with uniform retardation has a velocity of 40 ms⁻¹ at point a and 20 ms⁻¹ at another point b. if the particle comes to rest at a point c, 200 m from a, calculate the
i) distance ab;
ii) time taken to move from a to b;
iii) time taken to move from a to c.

list three physical quantities that can be deduced from a velocity - time graph

Question

a stone is projected vertically upward with a speed of 30 ms⁻¹ from the top of a tower of height 50 m. neglecting air resistance, determine g = 10 ms⁻²
a. the maximum height reached from the ground.
b. the time of flight
c. the total distance

question 2
a. sketch velocity - time graph to illustrate each of the following:
   i) uniform motion;
   ii) uniform acceleration with an initial velocity u = a m/s (a > 1)
   iii) a body moving with zero acceleration from point a to point b and then decelerate to point c
b. with an illustration how can we determine the
   i) the acceleration
   ii) deceleration
   iii) total distance

question 3
a. a particle moving in a straight line with uniform retardation has a velocity of 40 ms⁻¹ at point a and 20 ms⁻¹ at another point b. if the particle comes to rest at a point c, 200 m from a, calculate the
   i) distance ab;
   ii) time taken to move from a to b;
   iii) time taken to move from a to c.

list three physical quantities that can be deduced from a velocity - time graph

Accepted Answer

### Part 1 (Projectile from tower)
#### a. Maximum height from ground
# Explanation:
## Step 1: Find height above tower
Use $ v^2 = u^2 - 2gh_1 $ (upward motion, $ v = 0 $, $ u = 30 \, \text{m/s} $, $ g = 10 \, \text{m/s}^2 $)
$ 0 = 30^2 - 2(10)h_1 $
$ 20h_1 = 900 $
$ h_1 = 45 \, \text{m} $

## Step 2: Add tower height
Tower height $ h_2 = 50 \, \text{m} $, total height $ H = h_1 + h_2 $
$ H = 45 + 50 = 95 \, \text{m} $

# Answer: $ 95 \, \text{m} $

#### b. Time of flight
# Explanation:
## Step 1: Time to reach max height
Use $ v = u - gt_1 $ ( $ v = 0 $ )
$ 0 = 30 - 10t_1 $
$ t_1 = 3 \, \text{s} $

## Step 2: Time to fall from max height
Fall height $ H = 95 \, \text{m} $, use $ s = ut + \frac{1}{2}gt_2^2 $ ( $ u = 0 $ )
$ 95 = 0 + \frac{1}{2}(10)t_2^2 $
$ 5t_2^2 = 95 $
$ t_2^2 = 19 $
$ t_2 \approx 4.36 \, \text{s} $

## Step 3: Total time
$ T = t_1 + t_2 = 3 + 4.36 \approx 7.36 \, \text{s} $ (or use quadratic for downward displacement $ -50 = 30t - 5t^2 $, solve $ 5t^2 - 30t - 50 = 0 $, $ t = \frac{30 \pm \sqrt{900 + 1000}}{10} = \frac{30 \pm \sqrt{1900}}{10} \approx 7.36 \, \text{s} $)

# Answer: $ \approx 7.36 \, \text{s} $ (exact: $ 3 + \sqrt{19} \, \text{s} $)

#### c. Total distance
# Explanation:
## Step 1: Upward distance
Upward: $ 45 \, \text{m} $ (from tower top to max height)

## Step 2: Downward distance
Downward: $ 95 \, \text{m} $ (from max height to ground)

## Step 3: Sum distances
Total distance $ D = 45 + 95 = 140 \, \text{m} $

# Answer: $ 140 \, \text{m} $

### Question 2 (Velocity-Time Graphs)
#### a. i. Uniform motion
# Brief Explanations:
Uniform motion has constant velocity, so velocity-time graph is a **horizontal line** (parallel to time axis) at $ v = \text{constant} $.

# Answer: Horizontal line (e.g., $ v = 5 \, \text{m/s} $ for all $ t $)

#### a. ii. Uniform acceleration ( $ u = a \, \text{m/s}, a > 1 $ )
# Brief Explanations:
Uniform acceleration means velocity increases linearly. Graph is a **straight line with positive slope** (since $ a > 0 $), starting at $ v = a $ when $ t = 0 $, and slope $ = a $.

# Answer: Straight line with positive slope, $ v(0) = a $

#### a. iii. Zero acceleration (A to B) then deceleration (B to C)
# Brief Explanations:
- A to B: Zero acceleration → constant velocity (horizontal line).
- B to C: Deceleration → velocity decreases linearly (straight line with negative slope) until $ v = 0 $ (or some final velocity) at C.

# Answer: Horizontal line (A to B), then straight line with negative slope (B to C)

#### b. i. Acceleration from v-t graph
# Brief Explanations:
Acceleration is the **slope of the velocity-time graph** ( $ a = \frac{\Delta v}{\Delta t} $ ). For a straight line, slope $ = \frac{v_2 - v_1}{t_2 - t_1} $.

# Answer: Slope of v-t graph ( $ a = \frac{\Delta v}{\Delta t} $ )

#### b. ii. Deceleration from v-t graph
# Brief Explanations:
Deceleration is negative acceleration (or magnitude of slope when velocity decreases). If velocity decreases from $ v_1 $ to $ v_2 $ over $ \Delta t $, deceleration $ = \frac{v_1 - v_2}{\Delta t} $ (positive value, magnitude of negative slope).

# Answer: Magnitude of negative slope ( $ \frac{v_1 - v_2}{\Delta t} $ )

#### b. iii. Total distance from v-t graph
# Brief Explanations:
Total distance is the **area under the velocity-time graph** (integral of speed, but for v-t, area between graph and time axis, considering all intervals). For linear segments, calculate area of rectangles, triangles, trapezoids, etc., and sum.

# Answer: Area under v-t graph (sum of geometric areas)

### Question 3 (Uniform Retardation)
#### a. i. Distance AB
# Explanation:
Use $ v^2 = u^2 + 2as $ (retardation $ a = -r $, here $ u = 40 \, \text{m/s} $, $ v = 20 \, \text{m/s} $, find $ s = AB $)
$ 20^2 = 40^2 + 2(-r)s $ → Wait, retardation means $ a = -|a| $, but we can use $ v^2 = u^2 + 2as $ with $ a = -r $. Alternatively, since it’s uniform retardation, $ a $ is constant (negative).

Correct formula: $ v^2 = u^2 + 2as $ ( $ a $ is retardation, so $ a = -k $, but let's solve for $ s $)
$ 20^2 = 40^2 + 2a \cdot AB $
First, find $ a $ from AC: $ v = 0 $, $ u = 40 $, $ s = 200 \, \text{m} $
$ 0 = 40^2 + 2a(200) $
$ 400a = -1600 $
$ a = -4 \, \text{m/s}^2 $ (retardation $ 4 \, \text{m/s}^2 $)

Now use $ a = -4 $ for AB:
$ 20^2 = 40^2 + 2(-4) \cdot AB $
$ 400 = 1600 - 8 \cdot AB $
$ 8 \cdot AB = 1600 - 400 = 1200 $
$ AB = 150 \, \text{m} $

# Answer: $ 150 \, \text{m} $

#### a. ii. Time from A to B
# Explanation:
Use $ v = u + at $ ( $ u = 40 $, $ v = 20 $, $ a = -4 $ )
$ 20 = 40 + (-4)t $
$ -4t = -20 $
$ t = 5 \, \text{s} $

# Answer: $ 5 \, \text{s} $

#### a. iii. Time from A to C
# Explanation:
Use $ v = u + at $ ( $ v = 0 $, $ u = 40 $, $ a = -4 $ )
$ 0 = 40 + (-4)t $
$ -4t = -40 $
$ t = 10 \, \text{s} $

# Answer: $ 10 \, \text{s} $

#### b. Physical quantities from v-t graph
# Brief Explanations:
1. **Displacement**: Area under v-t graph (signed, for position change).
2. **Acceleration**: Slope of v-t graph ( $ \frac{\Delta v}{\Delta t} $ ).
3. **Instantaneous velocity**: Value of $ v $ at a specific $ t $ (y-axis value).

# Answer:
1. Displacement (area under graph)
2. Acceleration (slope)
3. Instantaneous velocity (v at a time)

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QUESTION IMAGE

Question

Explanation:

Part 1 (Projectile from tower)

a. Maximum height from ground

Step 1: Find height above tower

Step 2: Add tower height

Step 1: Time to reach max height

Step 2: Time to fall from max height

Step 3: Total time

Step 1: Upward distance

Step 2: Downward distance

Step 3: Sum distances

Answer:

b. Time of flight