what is the speed of the particle just before striking the ground when it is project with a velocity of 10m/s at angle 60degrees from the horizontal (angle of inclination = 30 degrees)

To determine the speed of a particle just before it strikes the ground when projected with an initial velocity of 10 m/s at an angle of 60 degrees from the horizontal, we can break down the problem into several steps. We will analyze the motion using the principles of projectile motion, which involves both horizontal and vertical components of the initial velocity.

Breaking Down the Initial Velocity

The initial velocity can be divided into horizontal and vertical components using trigonometric functions. Given that the initial velocity \( v_0 = 10 \, \text{m/s} \) and the angle of projection \( \theta = 60^\circ \), we can calculate the components as follows:

• Horizontal Component (v_x): \( v_x = v_0 \cdot \cos(\theta) = 10 \cdot \cos(60^\circ) = 10 \cdot 0.5 = 5 \, \text{m/s} \)
• Vertical Component (v_y): \( v_y = v_0 \cdot \sin(\theta) = 10 \cdot \sin(60^\circ) = 10 \cdot \frac{\sqrt{3}}{2} \approx 8.66 \, \text{m/s} \)

Analyzing Vertical Motion

As the particle moves upward and then downward, it will experience acceleration due to gravity, which is approximately \( g = 9.81 \, \text{m/s}^2 \). The vertical motion can be analyzed using the following kinematic equation to find the final vertical velocity just before impact:

Final Vertical Velocity (v_yf):

Using the equation: \( v_{yf}^2 = v_{y}^2 + 2g h \), where \( h \) is the maximum height reached. However, since we are interested in the speed just before impact, we can also use the time of flight to find the final vertical velocity.

Time of Flight Calculation

The time of flight \( T \) can be calculated using the formula:

\( T = \frac{2v_y}{g} = \frac{2 \cdot 8.66}{9.81} \approx 1.76 \, \text{s} \)

Final Vertical Velocity

Now, we can find the final vertical velocity just before hitting the ground:

\( v_{yf} = v_y - gT = 8.66 - 9.81 \cdot 1.76 \approx -8.66 \, \text{m/s} \)

Note that the negative sign indicates that the direction of the velocity is downward.

Calculating the Resultant Speed

To find the total speed just before impact, we combine the horizontal and vertical components of the velocity:

Using the Pythagorean theorem:

\( v_f = \sqrt{v_x^2 + v_{yf}^2} = \sqrt{(5)^2 + (-8.66)^2} \)

Calculating this gives:

\( v_f = \sqrt{25 + 75.0256} \approx \sqrt{100.0256} \approx 10 \, \text{m/s} \)

Final Thoughts

Thus, the speed of the particle just before it strikes the ground is approximately 10 m/s. This result illustrates the conservation of energy in projectile motion, where the initial kinetic energy is converted into kinetic energy just before impact, assuming no air resistance. The angle of inclination does not affect the final speed in this ideal scenario, as the speed remains constant throughout the flight, only changing direction.