A particle is projected upwards from the roof of a tower 60 m high with velocity `20 m//s.` Find
(a) the average speed and
(b) average velocity of the particle upto an instant when it strikes the ground. Take `g= 10 m//s^2`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the total time of flight (t)
The particle is projected upwards with an initial velocity (u) of 20 m/s from a height (h) of 60 m. We will use the equation of motion:

\[ s = ut + \frac{1}{2} a t^2 \]

Here,
- \( s \) is the total displacement (which will be -60 m since it falls to the ground),
- \( u = 20 \, \text{m/s} \) (upward),
- \( a = -g = -10 \, \text{m/s}^2 \) (downward).

Substituting these values into the equation, we get:

\[ -60 = 20t - \frac{1}{2} \cdot 10 t^2 \]

This simplifies to:

\[ -60 = 20t - 5t^2 \]

Rearranging gives us:

\[ 5t^2 - 20t - 60 = 0 \]

Dividing through by 5:

\[ t^2 - 4t - 12 = 0 \]

Now we can factor or use the quadratic formula to find \( t \):

\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1} \]

Calculating the discriminant:

\[ t = \frac{4 \pm \sqrt{16 + 48}}{2} = \frac{4 \pm \sqrt{64}}{2} = \frac{4 \pm 8}{2} \]

This gives us two possible solutions:

\[ t = \frac{12}{2} = 6 \, \text{s} \quad \text{(valid solution)} \]
\[ t = \frac{-4}{2} = -2 \, \text{s} \quad \text{(not valid)} \]

Thus, the total time of flight \( t = 6 \, \text{s} \).

### Step 2: Calculate the maximum height reached (h)
To find the maximum height reached by the particle, we can use the following equation:

\[ v^2 = u^2 + 2as \]

At the maximum height, the final velocity \( v = 0 \). Thus,

\[ 0 = (20)^2 + 2(-10)h \]

This simplifies to:

\[ 0 = 400 - 20h \]

Rearranging gives:

\[ 20h = 400 \implies h = 20 \, \text{m} \]

### Step 3: Calculate the total distance traveled
The total distance traveled by the particle includes:
1. The distance traveled upwards to the maximum height (20 m).
2. The distance traveled downwards from the maximum height to the ground (20 m) plus the height of the tower (60 m).

Thus, the total distance \( D \) is:

\[ D = 20 \, \text{m} + 60 \, \text{m} + 20 \, \text{m} = 100 \, \text{m} \]

### Step 4: Calculate the average speed
The average speed \( v_{avg} \) is given by:

\[ v_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{100 \, \text{m}}{6 \, \text{s}} \approx 16.67 \, \text{m/s} \]

### Step 5: Calculate the average velocity
The average velocity \( v_{avg, \text{vel}} \) is given by:

\[ v_{avg, \text{vel}} = \frac{\text{Total Displacement}}{\text{Total Time}} \]

The total displacement is the change in position from the starting point (top of the tower) to the ground:

\[ \text{Total Displacement} = -60 \, \text{m} \quad (\text{downward}) \]

Thus,

\[ v_{avg, \text{vel}} = \frac{-60 \, \text{m}}{6 \, \text{s}} = -10 \, \text{m/s} \]

### Final Answers:
(a) Average Speed = \( \approx 16.67 \, \text{m/s} \)
(b) Average Velocity = \( -10 \, \text{m/s} \)

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