Given:
- Mass of the body (m) = 3 kg
- Velocity of the body (v) = 10 m/s
- Angle of impact (θ) = 60 degrees
- Impact time (Δt) = 0.2 sec

To find:
- Impulsive force exerted on the wall

Explanation:

1. Impulse:
Impulse is defined as the change in momentum of an object. It can be calculated using the formula:

Impulse (J) = Change in momentum (Δp) = m * Δv

Where,
m = mass of the object
Δv = change in velocity of the object

2. Change in Momentum:
The change in momentum of the body can be calculated using the formula:

Δp = m * (v_f - v_i)

Where,
m = mass of the body
v_f = final velocity of the body after impact
v_i = initial velocity of the body before impact

3. Final Velocity:
To calculate the final velocity after impact, we need to resolve the initial velocity into horizontal and vertical components.

Horizontal component of velocity (v_x) = v * cos(θ)
Vertical component of velocity (v_y) = v * sin(θ)

Since the body returns at the same angle, the horizontal component of velocity remains the same, while the vertical component changes direction.

Final velocity in the horizontal direction (v_xf) = v_x = v * cos(θ)
Final velocity in the vertical direction (v_yf) = -v_y = -v * sin(θ)

4. Impulsive Force:
The impulsive force exerted on the wall can be calculated using the formula:

Impulsive force (F) = Impulse (J) / Impact time (Δt)

Calculation:
1. Change in momentum:
Δp = m * (v_f - v_i)
= 3 kg * ((v_xf - v_xi) i + (v_yf - v_yi) j)

2. Final velocities:
v_xf = v_x = v * cos(θ) = 10 m/s * cos(60°) = 5 m/s
v_yf = -v_y = -v * sin(θ) = -10 m/s * sin(60°) = -8.66 m/s

3. Change in momentum:
Δp = 3 kg * ((5 m/s - 10 m/s) i + (-8.66 m/s - 0 m/s) j)
= 3 kg * (-5 m/s i - 8.66 m/s j)
= -15 kg m/s i - 25.98 kg m/s j

4. Impulsive force:
F = Δp / Δt
= (-15 kg m/s i - 25.98 kg m/s j) / 0.2 s
= -75 kg m/s² i - 129.9 kg m/s² j

Answer:
The impulsive force exerted on the wall is -75 kg m/s² i - 129.9 kg m/s² j. The negative sign indicates that the force is in the opposite

The answer is 6√3..!!