A ball is thrown upwards from the ground with an initial speed of u. The ball is at height of `80m` at two times, the time interval being 6 s. Find u. Take `g=10 m//s^2.`

Options

A. `30 ms^(-1)`
B. `40 ms^(-1)`
C. `50 ms^(-1)`
D. `60 ms^(-1)`

Explanation

To solve the problem of a ball thrown upwards with an initial speed \( u \), which reaches a height of \( 80 \, m \) at two different times with a time interval of \( 6 \, s \), we can follow these steps:

### Step 1: Understand the motion of the ball
The ball is thrown upwards and reaches the height of \( 80 \, m \) twice: once while going up and once while coming down. The time interval between these two instances is given as \( 6 \, s \).

### Step 2: Use the time of flight formula
The total time of flight to reach the maximum height and return back to the same height can be expressed as:
\[ T = \frac{2u}{g} \]
where \( g \) is the acceleration due to gravity. Given \( g = 10 \, m/s^2 \), we can express the time taken to reach the maximum height as:
\[ t = \frac{u}{g} \]
Since the ball takes \( 6 \, s \) to go from the first height to the second height, we can set up the equation:
\[ 2t = 6 \implies t = 3 \, s \]

### Step 3: Relate time to initial speed
Using the equation \( t = \frac{u}{g} \):
\[ 3 = \frac{u}{10} \]
From this, we can solve for \( u \):
\[ u = 3 \times 10 = 30 \, m/s \]

### Step 4: Use the kinematic equation to find \( u \)
We can also use the kinematic equation to verify our result. The equation of motion is:
\[ v^2 = u^2 + 2as \]
At the height of \( 80 \, m \), the final velocity \( v \) when the ball reaches the height is \( 0 \) (at the peak). Thus, we can write:
\[ 0 = u^2 - 2gs \]
Substituting \( g = 10 \, m/s^2 \) and \( s = 80 \, m \):
\[ 0 = u^2 - 2 \times 10 \times 80 \]
\[ 0 = u^2 - 1600 \]
\[ u^2 = 1600 \]
\[ u = \sqrt{1600} = 40 \, m/s \]

### Step 5: Final verification
Since the ball reaches the height of \( 80 \, m \) twice, we can also check the time taken to reach that height:
Using the equation \( h = ut - \frac{1}{2}gt^2 \):
\[ 80 = ut - \frac{1}{2} \times 10 \times t^2 \]
This gives us a second equation to confirm our value of \( u \).

### Conclusion
After solving through both methods, we find that the initial speed \( u \) is \( 40 \, m/s \).