a projectile is fired at an angle of 30 to the horizontal such that the vertical component of its initial velocity is 80 m/s.its time of flight is T.its velocity at t=T/4 has a magnitude of nearly

To find the velocity of the projectile at \( t = \frac{T}{4} \), we first need to break down the problem into manageable parts. We know the vertical component of the initial velocity and the angle at which the projectile is fired. Let's go through the calculations step by step.

Understanding the Components of Motion

When a projectile is launched, its motion can be analyzed in two dimensions: horizontal and vertical. The initial vertical velocity (\( V_{y0} \)) is given as 80 m/s, and the launch angle is 30 degrees. The total time of flight (\( T \)) can be derived from the vertical motion equations.

Calculating Time of Flight

The time of flight for a projectile launched at an angle can be calculated using the formula:

• Time of Flight, \( T = \frac{2V_{y0}}{g} \)

Here, \( g \) is the acceleration due to gravity, approximately 9.81 m/s². Plugging in the values:

• \( T = \frac{2 \times 80 \text{ m/s}}{9.81 \text{ m/s}^2} \approx 16.3 \text{ seconds} \)

Finding Velocity at \( t = \frac{T}{4} \)

Next, we need to determine the velocity of the projectile at \( t = \frac{T}{4} \). Since \( T \) is approximately 16.3 seconds, we find:

• \( t = \frac{T}{4} = \frac{16.3}{4} \approx 4.075 \text{ seconds} \)

Now, we can calculate the vertical and horizontal components of the velocity at this time.

Vertical Velocity Component

The vertical velocity at any time \( t \) can be calculated using the equation:

• \( V_y = V_{y0} - gt \)

Substituting the values we have:

• \( V_y = 80 \text{ m/s} - (9.81 \text{ m/s}^2 \times 4.075 \text{ s}) \)
• \( V_y \approx 80 \text{ m/s} - 39.97 \text{ m/s} \approx 40.03 \text{ m/s} \)

Horizontal Velocity Component

The horizontal component of the velocity remains constant throughout the flight and can be calculated as:

• \( V_{x} = V_{0} \cos(\theta) \)

To find \( V_{0} \), we can use the vertical component:

• \( V_{0} = \frac{V_{y0}}{\sin(30^\circ)} = \frac{80 \text{ m/s}}{0.5} = 160 \text{ m/s} \)
• \( V_{x} = 160 \text{ m/s} \cos(30^\circ) = 160 \text{ m/s} \times \frac{\sqrt{3}}{2} \approx 138.56 \text{ m/s} \)

Calculating the Resultant Velocity

Now that we have both components of the velocity at \( t = \frac{T}{4} \), we can find the magnitude of the resultant velocity using the Pythagorean theorem:

• \( V = \sqrt{V_x^2 + V_y^2} \)

Substituting the values:

• \( V \approx \sqrt{(138.56 \text{ m/s})^2 + (40.03 \text{ m/s})^2} \)
• \( V \approx \sqrt{19200.43 + 1600.24} \approx \sqrt{20800.67} \approx 144.22 \text{ m/s} \)

Final Result

Thus, the magnitude of the velocity of the projectile at \( t = \frac{T}{4} \) is approximately 144.22 m/s. This value reflects the combination of both the horizontal and vertical components of motion at that specific time in the projectile's flight.