A ball is thrown vertically upwards with a velocity of `25 ms^(-1)` from the top of a tower of height 30m . How long will it travel before it hits ground

Options

A. 6s
B. 5sec
C. 4sec
D. 12s

Explanation

To solve the problem of how long the ball will take to hit the ground after being thrown upwards from the top of a tower, we can use the equations of motion. Here’s a step-by-step solution:

### Step 1: Identify the given data
- Initial velocity of the ball, \( u = 25 \, \text{m/s} \) (upwards)
- Height of the tower, \( h = 30 \, \text{m} \) (downwards)
- Acceleration due to gravity, \( g = 9.81 \, \text{m/s}^2 \) (downwards)

### Step 2: Define the displacement
Since the ball is thrown upwards and then falls down to the ground, the total displacement when it hits the ground will be the height of the tower, which is \( -30 \, \text{m} \) (negative because it is downward).

### Step 3: Use the equation of motion
We can use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \( s \) is the displacement,
- \( u \) is the initial velocity,
- \( a \) is the acceleration (which will be \( -g \) since it acts downward),
- \( t \) is the time.

Substituting the known values:
\[ -30 = 25t - \frac{1}{2} (9.81) t^2 \]

### Step 4: Rearranging the equation
Rearranging gives us:
\[ \frac{1}{2} (9.81) t^2 - 25t - 30 = 0 \]
Multiplying through by 2 to eliminate the fraction:
\[ 9.81 t^2 - 50t - 60 = 0 \]

### Step 5: Using the quadratic formula
This is a quadratic equation in the form \( at^2 + bt + c = 0 \), where:
- \( a = 9.81 \)
- \( b = -50 \)
- \( c = -60 \)

Using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting the values:
\[ t = \frac{-(-50) \pm \sqrt{(-50)^2 - 4 \cdot 9.81 \cdot (-60)}}{2 \cdot 9.81} \]
\[ t = \frac{50 \pm \sqrt{2500 + 2352.4}}{19.62} \]
\[ t = \frac{50 \pm \sqrt{4852.4}}{19.62} \]
Calculating the square root:
\[ t = \frac{50 \pm 69.7}{19.62} \]

### Step 6: Calculate the two possible values for \( t \)
Calculating the two possible values:
1. \( t = \frac{50 + 69.7}{19.62} \approx 6.1 \, \text{s} \)
2. \( t = \frac{50 - 69.7}{19.62} \) (this will yield a negative value which is not physically meaningful)

### Step 7: Conclusion
The only valid solution is:
\[ t \approx 6.1 \, \text{s} \]
Thus, the ball will take approximately **6.1 seconds** to hit the ground.

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