A body projected vertically upwards, reaches a height of 180 m. Find out `[g=10ms^(-2)]`
(i) intial velocity of body
(ii) time taken by body to reach its maximum hight
(iii) the time for which the body remains in air
(iv) height of the body from the ground after 8 sec.
(v) velocity of body at t=8 sec
(vi) the time at which the body reach the height of 100 m
(vii) A man at height of 135 m from ground tries to catch the body but he will be able to catch the body again ?

Answer

To solve the problem step by step, we will break down each part of the question regarding the body projected vertically upwards.

### Given Data:
- Maximum height (h) = 180 m
- Acceleration due to gravity (g) = 10 m/s²

### (i) Initial Velocity of the Body
To find the initial velocity (u), we can use the equation of motion:
\[ v^2 = u^2 + 2as \]
At maximum height, the final velocity (v) is 0. Thus, we have:
\[ 0 = u^2 - 2 \cdot g \cdot h \]
Substituting the values:
\[ 0 = u^2 - 2 \cdot 10 \cdot 180 \]
\[ u^2 = 3600 \]
\[ u = \sqrt{3600} = 60 \, \text{m/s} \]

### (ii) Time Taken to Reach Maximum Height
Using the equation:
\[ v = u + at \]
At maximum height, v = 0. Thus:
\[ 0 = 60 - 10t \]
Rearranging gives:
\[ 10t = 60 \]
\[ t = 6 \, \text{s} \]

### (iii) Total Time of Flight
The total time of flight (T) is the time to go up and come back down. Since the time to reach maximum height is 6 seconds, the time to fall back is also 6 seconds:
\[ T = 6 + 6 = 12 \, \text{s} \]

### (iv) Height of the Body from the Ground after 8 Seconds
Using the equation:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting the values:
\[ s = 60 \cdot 8 + \frac{1}{2} \cdot (-10) \cdot (8^2) \]
\[ s = 480 - 320 = 160 \, \text{m} \]

### (v) Velocity of the Body at t = 8 seconds
Using the equation:
\[ v = u + at \]
Substituting the values:
\[ v = 60 - 10 \cdot 8 \]
\[ v = 60 - 80 = -20 \, \text{m/s} \]
(The negative sign indicates the direction is downwards.)

### (vi) Time at which the Body Reaches the Height of 100 m
Using the equation:
\[ s = ut + \frac{1}{2} a t^2 \]
Setting s = 100 m:
\[ 100 = 60t - 5t^2 \]
Rearranging gives:
\[ 5t^2 - 60t + 100 = 0 \]
Dividing by 5:
\[ t^2 - 12t + 20 = 0 \]
Factoring:
\[ (t - 10)(t - 2) = 0 \]
Thus, \( t = 2 \, \text{s} \) and \( t = 10 \, \text{s} \).

### (vii) Will the Man Catch the Body?
The man is at a height of 135 m. Since the maximum height of the body is 180 m, the body will pass through 135 m on its way up and also on its way down. Therefore, yes, the man will be able to catch the body.

### Summary of Answers:
(i) Initial velocity = 60 m/s
(ii) Time to reach max height = 6 s
(iii) Total time in air = 12 s
(iv) Height after 8 s = 160 m
(v) Velocity at 8 s = -20 m/s
(vi) Time to reach 100 m = 2 s (and 10 s)
(vii) Yes, the man will catch the body.