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 of an object thrown vertically upwards that passes the height of 5 m twice in an interval of 10 seconds, we can follow these steps:

### Step 1: Understanding the Motion
The object is thrown upwards and will reach a maximum height before falling back down. The height of 5 m is crossed twice: once on the way up and once on the way down. The time interval between these two crossings is given as 10 seconds.

### Step 2: Time of Flight
The total time of flight (T) for an object thrown vertically upwards can be calculated using the time it takes to reach the maximum height and the time it takes to return to the ground. Since the object takes the same amount of time to go up as it does to come down, the time to reach the maximum height is T/2.

### Step 3: Analyzing the Given Information
Since the object passes the height of 5 m twice in 10 seconds, this means:
- The time taken to go from the ground to 5 m (upwards) is T1.
- The time taken to go from 5 m to the ground (downwards) is T2.
- We know that T2 - T1 = 10 seconds.

### Step 4: Using the Equation of Motion
Using the second equation of motion:
\[ s = ut + \frac{1}{2}gt^2 \]
where:
- \( s \) is the displacement (5 m),
- \( u \) is the initial velocity,
- \( g \) is the acceleration due to gravity (approximately 9.81 m/s²),
- \( t \) is the time taken to reach that height.

For the upward motion to reach 5 m:
\[ 5 = ut_1 - \frac{1}{2}gt_1^2 \]
For the downward motion from 5 m back to the ground:
\[ 5 = ut_2 - \frac{1}{2}gt_2^2 \]

### Step 5: Setting Up the Equations
From the first equation:
\[ ut_1 - \frac{1}{2}gt_1^2 = 5 \]
From the second equation:
\[ ut_2 - \frac{1}{2}gt_2^2 = 5 \]

### Step 6: Relating the Times
Since the time interval between the two heights is 10 seconds:
\[ t_2 - t_1 = 10 \]
We can express \( t_2 \) in terms of \( t_1 \):
\[ t_2 = t_1 + 10 \]

### Step 7: Solving the Equations
Substituting \( t_2 \) into the second equation:
\[ u(t_1 + 10) - \frac{1}{2}g(t_1 + 10)^2 = 5 \]

Now we have two equations with two unknowns (u and t1). We can solve these equations simultaneously to find the values of u and t1.

### Step 8: Finding Total Time of Flight
The total time of flight \( T \) is given by:
\[ T = t_1 + t_2 = t_1 + (t_1 + 10) = 2t_1 + 10 \]

### Conclusion
By solving the equations, we can find the value of \( t_1 \) and subsequently calculate the total time of flight \( T \).