A ball is projected upwards from the top of a tower with a velocity `50ms^-1` making an angle `30^@` with the horizontal. The height of tower is 70m. After how many seconds from the instant of throwing, will the ball reach the ground. `(g=10 ms^-2)`

Options

A. 2s
B. 5s
C. 7s
D. 9s

Explanation

To solve the problem of a ball projected upwards from the top of a tower, we will follow these steps:

### Step 1: Identify the given values
- Initial velocity (u) = 50 m/s
- Angle of projection (θ) = 30°
- Height of the tower (h) = 70 m
- Acceleration due to gravity (g) = 10 m/s²

### Step 2: Resolve the initial velocity into components
The initial velocity can be resolved into horizontal (u_x) and vertical (u_y) components:
- \( u_y = u \sin(\theta) \)
- \( u_x = u \cos(\theta) \)

Calculating \( u_y \):
\[ u_y = 50 \sin(30°) = 50 \times \frac{1}{2} = 25 \text{ m/s} \]

### Step 3: Set up the equation of motion
We will use the equation of motion to find the time taken to reach the ground. The vertical displacement (s_y) is negative because the ball is falling down from the height of the tower:
\[ s_y = -70 \text{ m} \]

Using the equation of motion:
\[ s_y = u_y t + \frac{1}{2} (-g) t^2 \]
Substituting the known values:
\[ -70 = 25t - \frac{1}{2} \cdot 10 t^2 \]
This simplifies to:
\[ -70 = 25t - 5t^2 \]

### Step 4: Rearrange the equation
Rearranging gives:
\[ 5t^2 - 25t - 70 = 0 \]
Dividing the entire equation by 5:
\[ t^2 - 5t - 14 = 0 \]

### Step 5: Solve the quadratic equation
We can solve the quadratic equation using the factorization method:
\[ t^2 - 7t + 2t - 14 = 0 \]
Factoring gives:
\[ (t + 2)(t - 7) = 0 \]
Setting each factor to zero gives:
\[ t + 2 = 0 \quad \Rightarrow \quad t = -2 \quad \text{(not valid)} \]
\[ t - 7 = 0 \quad \Rightarrow \quad t = 7 \text{ seconds} \]

### Conclusion
The time taken for the ball to reach the ground is **7 seconds**.

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