A particle is projected with velocity u at an angle of ? from (0,1). State whether the particle will cross the coordinate (4,9)after 1s ?

To determine whether a particle projected from the point (0, 1) with an initial velocity \( u \) at an angle \( \theta \) will cross the coordinate (4, 9) after 1 second, we need to analyze the motion of the particle using the equations of projectile motion.

Understanding Projectile Motion

Projectile motion can be broken down into horizontal and vertical components. The initial velocity \( u \) can be split into two components:

• Horizontal component: \( u_x = u \cos(\theta) \)
• Vertical component: \( u_y = u \sin(\theta) \)

Position After 1 Second

After \( t \) seconds, the horizontal and vertical positions of the particle can be calculated using the following equations:

• Horizontal position: \( x(t) = u_x t = u \cos(\theta) t \)
• Vertical position: \( y(t) = y_0 + u_y t - \frac{1}{2} g t^2 \)

Here, \( y_0 \) is the initial vertical position (which is 1 in this case), and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

Calculating Positions at \( t = 1 \) Second

Substituting \( t = 1 \) second into the equations:

• Horizontal position: \( x(1) = u \cos(\theta) \)
• Vertical position: \( y(1) = 1 + u \sin(\theta) - \frac{1}{2} g (1)^2 \)

Now, we need to check if these positions equal (4, 9) after 1 second.

Setting Up the Equations

For the particle to cross the coordinate (4, 9), we set up the following equations:

• From the horizontal position: \( u \cos(\theta) = 4 \)
• From the vertical position: \( 1 + u \sin(\theta) - \frac{1}{2} g = 9 \)

Solving the Vertical Position Equation

Rearranging the vertical position equation gives:

\( u \sin(\theta) = 9 - 1 + \frac{1}{2} g = 8 + 4.905 = 12.905 \)

Finding Relationships Between Components

Now we have two equations:

• 1. \( u \cos(\theta) = 4 \)
• 2. \( u \sin(\theta) = 12.905 \)

To find \( u \), we can use the Pythagorean identity:

\( u^2 = (u \cos(\theta))^2 + (u \sin(\theta))^2 \)

Substituting the values:

\( u^2 = 4^2 + (12.905)^2 \)

Calculating gives:

\( u^2 = 16 + 166.5 = 182.5 \)

Thus, \( u = \sqrt{182.5} \approx 13.5 \, \text{m/s} \).

Verifying the Angles

Now we can find \( \theta \) using:

• \( \tan(\theta) = \frac{u \sin(\theta)}{u \cos(\theta)} = \frac{12.905}{4} \)

Calculating gives:

\( \tan(\theta) \approx 3.226 \), which means \( \theta \approx 72.3^\circ \).

Conclusion

With the calculated values of \( u \) and \( \theta \), we find that the particle can indeed reach the point (4, 9) after 1 second, provided it is projected with the appropriate initial velocity and angle. Therefore, yes, the particle will cross the coordinate (4, 9) after 1 second.