A ball is projected with a velocity of `96 ft//"sec."` (i) its max, range on the horizontal plane (ii) the direction of projection for range of 144 ft. are

Options

A. `288, 30^(@), 60^(@)`
B. `100, 45^(@), 45^(@)`
C. `288, 15^(@), 75^(@)`
D. None

Answer

To solve the problem step by step, we will follow the physics of projectile motion.

### Step 1: Calculate the Maximum Range
The formula for the maximum range \( R \) of a projectile launched at an angle \( \theta \) with an initial velocity \( U \) is given by:

\[ R = \frac{U^2 \sin 2\theta}{g} \]

For maximum range, the angle \( \theta \) is \( 45^\circ \). Therefore, \( \sin 2\theta = \sin 90^\circ = 1 \).

Given:
- \( U = 96 \, \text{ft/sec} \)
- \( g \) (acceleration due to gravity) is approximately \( 32 \, \text{ft/sec}^2 \) (standard value for feet).

Substituting the values into the formula:

\[ R = \frac{(96)^2 \cdot 1}{32} \]

Calculating \( R \):

\[ R = \frac{9216}{32} = 288 \, \text{ft} \]

### Step 2: Determine the Direction of Projection for a Range of 144 ft
Now, we need to find the angle \( \theta \) for which the range \( R \) is \( 144 \, \text{ft} \).

Using the range formula again:

\[ R = \frac{U^2 \sin 2\theta}{g} \]

Setting \( R = 144 \, \text{ft} \):

\[ 144 = \frac{(96)^2 \sin 2\theta}{32} \]

Rearranging gives:

\[ \sin 2\theta = \frac{144 \cdot 32}{9216} \]

Calculating the right side:

\[ \sin 2\theta = \frac{4608}{9216} = \frac{1}{2} \]

### Step 3: Solve for \( 2\theta \)
Now, we find \( 2\theta \):

\[ 2\theta = \sin^{-1}\left(\frac{1}{2}\right) \]

This gives:

\[ 2\theta = 30^\circ \]

Thus, dividing by 2 gives:

\[ \theta = 15^\circ \]

### Step 4: Determine the Complementary Angle
Since the range is the same for complementary angles, the other angle is:

\[ \theta' = 90^\circ - 15^\circ = 75^\circ \]

### Final Answers
The maximum range on the horizontal plane is \( 288 \, \text{ft} \) and the directions of projection for a range of \( 144 \, \text{ft} \) are \( 15^\circ \) and \( 75^\circ \).