from the top of multi-storeyed building ,39.2 m tall , a boy projects a stone vertically upward with an initial velocity of `9.8 ms^(-1)` such that it finally drops to the ground (i) when will the stone reach the ground? (ii) whn will it pass through the point of projection ? (iii) what will be its velocity before striking the ground ? take g =`10ms^(-2)`

Answer

To solve the problem, we will break it down into three parts as per the questions asked:

### Given Data:
- Height of the building (h) = 39.2 m
- Initial velocity of the stone (u) = 9.8 m/s (upward)
- Acceleration due to gravity (g) = 10 m/s² (downward)

### (i) When will the stone reach the ground?

We can use the second equation of motion:
\[ S = ut + \frac{1}{2} a t^2 \]

Here, we consider downward direction as positive. Therefore:
- Displacement (S) = 39.2 m (downward)
- Initial velocity (u) = -9.8 m/s (upward, hence negative)
- Acceleration (a) = 10 m/s² (downward)

Substituting these values into the equation:
\[ 39.2 = -9.8t + \frac{1}{2}(10)t^2 \]
\[ 39.2 = -9.8t + 5t^2 \]
Rearranging gives:
\[ 5t^2 - 9.8t - 39.2 = 0 \]

This is a quadratic equation in the form of \( at^2 + bt + c = 0 \), where:
- \( a = 5 \)
- \( b = -9.8 \)
- \( c = -39.2 \)

Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ t = \frac{-(-9.8) \pm \sqrt{(-9.8)^2 - 4 \cdot 5 \cdot (-39.2)}}{2 \cdot 5} \]
\[ t = \frac{9.8 \pm \sqrt{96.04 + 784}}{10} \]
\[ t = \frac{9.8 \pm \sqrt{880.04}}{10} \]
\[ t = \frac{9.8 \pm 29.66}{10} \]

Calculating the two possible values for \( t \):
1. \( t = \frac{9.8 + 29.66}{10} = \frac{39.46}{10} = 3.946 \) seconds (valid)
2. \( t = \frac{9.8 - 29.66}{10} = \frac{-19.86}{10} \) (not valid since time cannot be negative)

Thus, the time when the stone reaches the ground is approximately:
\[ t \approx 3.95 \text{ seconds} \]

### (ii) When will it pass through the point of projection?

To find when the stone passes through the point of projection, we first need to determine the time it takes to reach the maximum height. At maximum height, the final velocity (v) is 0.

Using the first equation of motion:
\[ v = u - gt \]
Setting \( v = 0 \):
\[ 0 = 9.8 - 10t \]
\[ 10t = 9.8 \]
\[ t = 0.98 \text{ seconds} \]

The stone will take the same time to come back down to the point of projection:
Total time to return to the point of projection:
\[ t_{total} = 2 \times 0.98 = 1.96 \text{ seconds} \]

### (iii) What will be its velocity before striking the ground?

Using the first equation of motion again:
\[ v = u - gt \]
Substituting the known values:
\[ v = 9.8 - 10 \times 3.95 \]
\[ v = 9.8 - 39.5 \]
\[ v = -29.7 \text{ m/s} \]

The negative sign indicates that the velocity is directed downward. Thus, the velocity before striking the ground is:
\[ v \approx 29.7 \text{ m/s downward} \]

### Summary of Answers:
1. Time to reach the ground: **3.95 seconds**
2. Time to pass through the point of projection: **1.96 seconds**
3. Velocity before striking the ground: **29.7 m/s downward**