A projectile has initially the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of `3 m s^-2` for `0.5 min`. If the maximum height reached by it is `80 m`, then the angle of projection is `(g = 10 ms^-2)`.

Options

A. `tan^-1 3`
B. `tan^-1(3//2)`
C. `tan^-1 (4//9)`
D. `sin^-1(4//9)`

Explanation

To solve the problem step by step, we will follow the given information and apply the relevant physics concepts.

### Step 1: Calculate the horizontal velocity of the projectile
The projectile has the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of \(3 \, \text{m/s}^2\) for \(0.5 \, \text{min}\).

First, convert \(0.5 \, \text{min}\) to seconds:
\[ 0.5 \, \text{min} = 0.5 \times 60 = 30 \, \text{s} \]

Now, use the formula for velocity under uniform acceleration:
\[ v = u + at \]
Since the initial velocity \(u = 0\), the formula simplifies to:
\[ v = at = 3 \, \text{m/s}^2 \times 30 \, \text{s} = 90 \, \text{m/s} \]

Thus, the horizontal component of the projectile's initial velocity is:
\[ u \cos \theta = 90 \, \text{m/s} \]

### Step 2: Relate the maximum height to the vertical component of velocity
The maximum height \(H\) reached by the projectile is given as \(80 \, \text{m}\). The formula for maximum height in projectile motion is:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
Substituting the known values, where \(g = 10 \, \text{m/s}^2\):
\[ 80 = \frac{u^2 \sin^2 \theta}{2 \times 10} \]
This simplifies to:
\[ 80 = \frac{u^2 \sin^2 \theta}{20} \]
Multiplying both sides by \(20\):
\[ 1600 = u^2 \sin^2 \theta \]

### Step 3: Express \(u \sin \theta\) in terms of \(u \cos \theta\)
From the previous step, we have:
\[ u \sin \theta = \sqrt{1600} = 40 \, \text{m/s} \]

### Step 4: Find the ratio of vertical and horizontal components
Now we have:
1. \(u \cos \theta = 90 \, \text{m/s}\)
2. \(u \sin \theta = 40 \, \text{m/s}\)

We can find the tangent of the angle of projection:
\[ \tan \theta = \frac{u \sin \theta}{u \cos \theta} = \frac{40}{90} = \frac{4}{9} \]

### Step 5: Calculate the angle of projection
To find the angle \(\theta\), we take the inverse tangent:
\[ \theta = \tan^{-1}\left(\frac{4}{9}\right) \]

### Final Answer
Thus, the angle of projection is:
\[ \theta = \tan^{-1}\left(\frac{4}{9}\right) \]