A stone of mass 0.50 kg is dropped from a height of 20 m above the ground. Calculate its speed just before it hits the ground. (g = 10 m s -2, air resistance is negligible.)

Step 1: Identify the given values for the stone's motion. The stone is dropped, so its initial velocity u = 0 m/s. The height h = 20 m. The acceleration due to gravity g = 10 m/s^2. We need to find the final speed v. Step 2: Use the kinematic equation that relates initial velocity, final velocity, acceleration, and displacement. Since air resistance is negligible, we can use the equation: v^2 = u^2 + 2gh Step 3: Substitute the values into the equation and solve for v. v^2 = (0 m/s)^2 + 2(10 m/s^2)(20 m) v^2 = 0 + 400 m^2/s^2 v = sqrt(400 m)^2/s^2 v = 20 m/s The speed of the stone just before it hits the ground is 20 m/s. 7. Step 1: Identify the given values for the lens problem. The object distance u = 30 cm. The focal length of the converging lens f = 15 cm. We need to find the image distance v. Step 2: Use the lens formula. The lens formula is given by: (1)/(f) = (1)/(u) + (1)/(v) Step 3: Rearrange the formula to solve for v. (1)/(v) = (1)/(f) - (1)/(u) Step 4: Substitute the given values into the formula and calculate v. (1)/(v) = (1)/(15 cm) - (1)/(30 cm) To subtract the fractions, find a common denominator, which is 30 cm. (1)/(v) = (2)/(30 cm) - (1)/(30 cm) (1)/(v) = (2 - 1)/(30 cm) (1)/(v) = (1)/(30 cm) v = 30 cm The position of the image formed is 30 cm. Since v is positive, the image is real and formed on the opposite side of the lens from the object. That's 2 down. 3 left today — send the next one.