Time of flight is 1 s and range is 4 m .Find the projection speed is:

Options

A. `sqrt(37)ms^(-1)`
B. `sqrt(41)ms^(-1)`
C. `sqrt(14)ms^(-1)`
D. `sqrt(40)ms^(-1)`

Explanation

To find the projection speed given the time of flight and range, we can follow these steps:

### Step 1: Understand the given information
- Time of flight (T) = 1 s
- Range (R) = 4 m
- Acceleration due to gravity (g) = 10 m/s² (approximately)

### Step 2: Use the formula for time of flight
The formula for time of flight in projectile motion is given by:
\[ T = \frac{2u \sin \theta}{g} \]
Substituting the known values:
\[ 1 = \frac{2u \sin \theta}{10} \]
Rearranging gives:
\[ u \sin \theta = \frac{g}{2} = \frac{10}{2} = 5 \text{ m/s} \]

### Step 3: Use the formula for range
The formula for range in projectile motion is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Substituting the known values:
\[ 4 = \frac{u^2 \sin 2\theta}{10} \]
Rearranging gives:
\[ u^2 \sin 2\theta = 40 \]

### Step 4: Relate sin 2θ to sin θ and cos θ
Using the identity \(\sin 2\theta = 2 \sin \theta \cos \theta\), we can rewrite the range equation:
\[ u^2 (2 \sin \theta \cos \theta) = 40 \]
This simplifies to:
\[ u^2 \sin \theta \cos \theta = 20 \]

### Step 5: Solve for u
We have two equations now:
1. \(u \sin \theta = 5\)
2. \(u^2 \sin \theta \cos \theta = 20\)

From the first equation, we can express \(\sin \theta\) in terms of \(u\):
\[ \sin \theta = \frac{5}{u} \]

Substituting this into the second equation:
\[ u^2 \left(\frac{5}{u}\right) \cos \theta = 20 \]
This simplifies to:
\[ 5u \cos \theta = 20 \]
Thus, we have:
\[ u \cos \theta = 4 \text{ m/s} \]

### Step 6: Find the projection speed
Now we have:
- \(u \sin \theta = 5\)
- \(u \cos \theta = 4\)

Using the Pythagorean theorem:
\[ u = \sqrt{(u \sin \theta)^2 + (u \cos \theta)^2} \]
Substituting the values:
\[ u = \sqrt{5^2 + 4^2} = \sqrt{25 + 16} = \sqrt{41} \]

### Final Answer
The projection speed \(u\) is:
\[ u = \sqrt{41} \text{ m/s} \]

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