a particle of mass 2.00m is projected at an angle of 45 degrees with the horizontal with a speed of 20root2 m/s.after 1 s an explosion occurs and the particle is broken into two equal parts.one piece is momentarily at rest before it falls.find the maximum height attained by the other piece.

To solve this problem, we need to analyze the motion of the particle before and after the explosion. Let's break it down step by step, starting with the initial conditions of the particle and then examining the effects of the explosion.

Initial Motion of the Particle

The particle is projected at an angle of 45 degrees with an initial speed of \(20\sqrt{2}\) m/s. We can find the horizontal and vertical components of the initial velocity using trigonometric functions.

• Horizontal Component (Vx):

  Vx = V * cos(θ) = \(20\sqrt{2} \cdot \cos(45^\circ) = 20\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 20\) m/s

• Vertical Component (Vy):

  Vy = V * sin(θ) = \(20\sqrt{2} \cdot \sin(45^\circ) = 20\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 20\) m/s

Position After 1 Second

Next, we need to determine the position of the particle after 1 second. We can use the equations of motion to find the horizontal and vertical positions.

• Horizontal Position (x):

  x = Vx * t = \(20 \cdot 1 = 20\) m

• Vertical Position (y):

  y = Vy * t - \(\frac{1}{2} g t^2\) = \(20 \cdot 1 - \frac{1}{2} \cdot 9.81 \cdot 1^2 = 20 - 4.905 = 15.095\) m

Explosion and Its Aftermath

At the end of 1 second, the particle undergoes an explosion, breaking into two equal parts. One piece comes to rest momentarily, which means it has no velocity at that instant. The other piece will continue moving with the velocity of the original particle just before the explosion.

Velocity Just Before the Explosion

Before the explosion, the vertical and horizontal components of the velocity remain unchanged (ignoring air resistance). Therefore, the velocity of the moving piece just before the explosion is:

• Horizontal Velocity (Vx): 20 m/s
• Vertical Velocity (Vy): 20 m/s

Calculating the Maximum Height of the Moving Piece

After the explosion, the moving piece will continue to ascend until it reaches its maximum height. At this point, its vertical velocity will become zero. We can use the following kinematic equation to find the maximum height:

Using the formula: \[ v^2 = u^2 + 2as \] where:

• v = final velocity (0 m/s at maximum height)
• u = initial vertical velocity (20 m/s)
• a = acceleration due to gravity (-9.81 m/s²)
• s = maximum height (h)

Setting up the equation:

0 = \(20^2 + 2 \cdot (-9.81) \cdot h\)

0 = 400 - 19.62h

19.62h = 400

h = \(\frac{400}{19.62} \approx 20.39\) m

Final Height Calculation

The maximum height attained by the moving piece is the height it reached after the explosion plus the height it was at just before the explosion:

Maximum height = height at 1 second + height gained after explosion

Maximum height = 15.095 m + 20.39 m = 35.485 m

Thus, the maximum height attained by the other piece after the explosion is approximately 35.49 meters.