A projectile is thrown with a velocity of `10sqrt(2)m//s` at an angle of `45^(@)` with horizontal. The interval between the moments when speed is `sqrt(125)m//s` is `(g=10m//s^(2))`

Options

A. `1.0s`
B. `1.5s`
C. `2.0s`
D. `0.5s`

Answer

To solve the problem step-by-step, we will analyze the projectile motion of the object thrown with a given initial velocity and angle.

### Step 1: Determine the initial velocity components
The projectile is thrown with a velocity of \(10\sqrt{2} \, \text{m/s}\) at an angle of \(45^\circ\). We can find the horizontal and vertical components of the initial velocity (\(u_x\) and \(u_y\)) using trigonometric functions:

\[ u_x = u \cos \theta = 10\sqrt{2} \cos 45^\circ = 10 \, \text{m/s} \]
\[ u_y = u \sin \theta = 10\sqrt{2} \sin 45^\circ = 10 \, \text{m/s} \]

### Step 2: Use the speed condition
The speed of the projectile at two different moments is given as \(\sqrt{125} \, \text{m/s}\). We can express the speed in terms of its components:

\[ v = \sqrt{u_x^2 + v_y^2} \]

Setting this equal to \(\sqrt{125}\):

\[ \sqrt{10^2 + v_y^2} = \sqrt{125} \]

Squaring both sides:

\[ 100 + v_y^2 = 125 \]

Solving for \(v_y^2\):

\[ v_y^2 = 125 - 100 = 25 \]

Thus, we have:

\[ v_y = \pm 5 \, \text{m/s} \]

### Step 3: Find times corresponding to \(v_y = 5 \, \text{m/s}\) and \(v_y = -5 \, \text{m/s}\)
Using the equation of motion for vertical velocity:

\[ v_y = u_y - g t \]

1. For \(v_y = 5 \, \text{m/s}\):

\[ 5 = 10 - 10t \implies 10t = 10 - 5 \implies t = \frac{1}{2} \, \text{s} \quad (t_1) \]

2. For \(v_y = -5 \, \text{m/s}\):

\[ -5 = 10 - 10t \implies 10t = 10 + 5 \implies t = \frac{15}{10} = \frac{3}{2} \, \text{s} \quad (t_2) \]

### Step 4: Calculate the time interval
The time interval between the two moments when the speed is \(\sqrt{125} \, \text{m/s}\) is:

\[ \Delta t = t_2 - t_1 = \frac{3}{2} - \frac{1}{2} = 1 \, \text{s} \]

### Final Answer
The interval between the moments when the speed is \(\sqrt{125} \, \text{m/s}\) is \(1 \, \text{s}\).

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