Step 1: Impulse and Change in Momentum
The impulse imparted to the ball during its collision with the ground is equal to the change in its momentum: J = Īp = m (v_f - v_i), where m is the mass of the ball, v_f is the final velocity, and v_i is the initial velocity.
Step 2: Find the Velocity of the Ball Just Before Hitting the Ground
Using the equation of motion vĀ² = uĀ² + 2as, where u = 0, a = g = 9.8 m/sĀ², and s = 10 m, we get: v_iĀ² = 0Ā² + 2(9.8)(10) = 196, v_i = ā196 = 14 m/s (downwards, so v_i = -14j).
Step 3: Find the Velocity of the Ball Just After Hitting the Ground
The ball rises to a height of 1.5 m. At the highest point, the velocity is 0. Using the equation of motion vĀ² = uĀ² + 2as, where v = 0, a = -g = -9.8 m/sĀ², and s = 1.5 m, we get: 0Ā² = v_fĀ² + 2(-9.8)(1.5), v_fĀ² = 29.4, v_f = ā29.4 ā 5.42 m/s (upwards, so v_f = 5.42j).
Step 4: Calculate the Impulse
Now, let's calculate the impulse: J = m (v_f - v_i) = 0.5 (5.42j - (-14j)) = 0.5 (5.42 + 14)j, J = 0.5 Ć 19.42j = 9.71j Ns.
Conclusion:
The impulse imparted to the ball is approximately 9.71 Ns. Thus, the final answer is 9.71 Ns.