An object thrown verticallly up from the ground passes the height 5 m twice in an interval of 10 s. What is its time of flight?

Options

A. `sqrt(28)s`
B. `sqrt(86)s`
C. `sqrt(104)s`
D. `sqrt(72)s`

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript.

### Step 1: Understand the problem
An object is thrown vertically upward and passes the height of 5 m twice in an interval of 10 seconds. We need to find the total time of flight of the object.

### Step 2: Analyze the motion
When the object is thrown upwards, it will reach a maximum height and then come back down. The object will pass the height of 5 m twice: once while going up and once while coming down.

### Step 3: Use the time interval
The time taken to pass the height of 5 m twice is given as 10 seconds. This means the time taken to go from the first 5 m point to the second 5 m point (at the same height) is 10 seconds.

### Step 4: Relate time of flight to initial velocity
The total time of flight (T) can be expressed as:
\[ T = 2u/g \]
where \( u \) is the initial velocity and \( g \) is the acceleration due to gravity (approximately 10 m/s²).

### Step 5: Use the kinematic equation
To find the initial velocity \( u \), we can use the kinematic equation for the height of 5 m:
\[ v^2 = u^2 - 2gh \]
At the height of 5 m, we can express the final velocity \( v \) when the object is at that height while moving upwards and downwards.

### Step 6: Calculate the velocity at 5 m
Using the kinematic equation:
1. For the upward motion at height 5 m:
\[ v^2 = u^2 - 2g(5) \]
\[ v^2 = u^2 - 100 \]

2. For the downward motion at height 5 m:
The velocity just before reaching the height again is the same magnitude but opposite in direction, so we can use the same equation.

### Step 7: Use displacement equation
Since the total displacement for the entire flight is zero (it returns to the ground), we can use the displacement equation:
\[ s = ut + \frac{1}{2}(-g)t^2 \]
Setting \( s = 0 \) for the total time of flight:
\[ 0 = u(10) - \frac{1}{2}(10)(10^2) \]
\[ 0 = 10u - 500 \]
\[ 10u = 500 \]
\[ u = 50 \, \text{m/s} \]

### Step 8: Calculate the time of flight
Now substitute \( u \) back into the time of flight equation:
\[ T = \frac{2u}{g} = \frac{2(50)}{10} = 10 \, \text{s} \]

### Final Answer
The total time of flight is **10 seconds**.

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