A particle falls freely from rest through a distance d. Its speed is then; A coconut falls from a 12 m tree. What would be its velocity half-way down?

Here are the solutions for Question 1 and Question 2. Question 1: A particle falls freely from rest through a distance d. Its speed is then; Step 1: Identify the given information and the relevant kinematic equation. The particle falls from rest, so the initial velocity u = 0. The distance fallen is s = d. The acceleration due to gravity is a = g. We need to find the final speed v. The kinematic equation relating these variables is v^2 = u^2 + 2as. Step 2: Substitute the values into the equation. v^2 = (0)^2 + 2(g)(d) v^2 = 2gd Step 3: Solve for v. Since speed is a scalar quantity and always positive, we take the positive square root. v = sqrt(2gd) Comparing this with the given options, the correct option is B. The final answer is B. sqrt(2gd). Question 2: A coconut falls from a 12 m tree. What would be its velocity half-way down? Step 1: Identify the given information. Total height of the tree H = 12 \, m. The coconut falls from rest, so initial velocity u = 0. We need to find the velocity when it is half-way down. Distance fallen s = (H)/(2) = 12 \, m2 = 6 \, m. Acceleration due to gravity g = 9.8 \, m/s^2. Step 2: Use the kinematic equation v^2 = u^2 + 2as. Substitute the values: v^2 = (0)^2 + 2(9.8 \, m/s^2)(6 \, m) Step 3: Calculate the square of the velocity. v^2 = 2 × 9.8 × 6 \, m^2/s^2 v^2 = 117.6 \, m^2/s^2 Step 4: Calculate the velocity. v = sqrt(117.6) \, m/s v ≈ 10.84 \, m/s Step 5: Compare the calculated velocity with the given options. The calculated velocity 10.84 \, m/s is closest to 11 \, m/s. The final answer is B. 11 \, m/s. That's 2 down. 3 left today — send the next one.