A stone is projected so that its horizontal range is maximum and equal to 80 ft. Its time of flight and the height it rises are

Options

A. `sqrt(3), 1`
B. `sqrt(4), 15`
C. `sqrt(5), 20`
D. None

Answer

To solve the problem step by step, we will first determine the time of flight and then the maximum height reached by the stone when it is projected at an angle that gives the maximum horizontal range.

### Step 1: Determine the angle of projection
For maximum horizontal range, the angle of projection (θ) is 45 degrees.

### Step 2: Use the formula for maximum range
The formula for the maximum range (R) of a projectile is given by:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
Since θ = 45 degrees, we have:
\[ \sin(2\theta) = \sin(90^\circ) = 1 \]
Thus, the formula simplifies to:
\[ R = \frac{u^2}{g} \]

### Step 3: Substitute the given range
Given that the maximum range (R) is 80 ft, we can set up the equation:
\[ 80 = \frac{u^2}{g} \]

### Step 4: Use the value of g
Assuming g (acceleration due to gravity) is approximately 32 ft/s² (standard value in feet), we can rearrange the equation to find u²:
\[ u^2 = R \cdot g = 80 \cdot 32 = 2560 \]

### Step 5: Calculate the initial velocity (u)
Now, we take the square root to find u:
\[ u = \sqrt{2560} \approx 50.6 \text{ ft/s} \]

### Step 6: Calculate the time of flight (T)
The time of flight (T) for a projectile is given by:
\[ T = \frac{2u \sin(\theta)}{g} \]
Substituting θ = 45 degrees:
\[ \sin(45^\circ) = \frac{\sqrt{2}}{2} \]
Thus, the formula becomes:
\[ T = \frac{2u \cdot \frac{\sqrt{2}}{2}}{g} = \frac{u \sqrt{2}}{g} \]

Now substituting the values:
\[ T = \frac{50.6 \cdot \sqrt{2}}{32} \approx \frac{50.6 \cdot 1.414}{32} \approx \frac{71.5}{32} \approx 2.23 \text{ seconds} \]

### Step 7: Calculate the maximum height (H)
The maximum height (H) can be calculated using the formula:
\[ H = \frac{u^2 \sin^2(\theta)}{2g} \]
Substituting θ = 45 degrees:
\[ \sin^2(45^\circ) = \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{2} \]
Thus, the formula becomes:
\[ H = \frac{u^2 \cdot \frac{1}{2}}{2g} = \frac{u^2}{4g} \]

Substituting the values:
\[ H = \frac{2560}{4 \cdot 32} = \frac{2560}{128} = 20 \text{ ft} \]

### Final Answers
- Time of Flight (T) ≈ 2.23 seconds
- Maximum Height (H) = 20 ft