Question

A ball of mass 10g is dropped from a height of 50m. If 30J energy is lost due to air resistance, then what is the velocity of the ball when it hits the ground?

• A. \(20 \, \text{m/s}\)
• B. \(25 \, \text{m/s}\)
• C. \(30 \, \text{m/s}\)
• D. \(35 \, \text{m/s}\)

Hint:

When calculating the velocity of a falling object, account for the energy lost due to air resistance and apply the conservation of energy principle.

Answer 1

The Correct Option is B

The total mechanical energy (potential + kinetic) of the ball is conserved except for the energy lost due to air resistance. The potential energy \(PE\) at height \(h\) is given by: \[ PE = mgh \] Where: - \(m = 10 \, \text{g} = 0.01 \, \text{kg}\), - \(g = 10 \, \text{m/s}^2\), - \(h = 50 \, \text{m}\). Thus, \[ PE = 0.01 \times 10 \times 50 = 5 \, \text{J} \] The energy lost due to air resistance is 30J, so the remaining energy after air resistance is: \[ KE = 5 \, \text{J} - 30 \, \text{J} = 25 \, \text{J} \] The final kinetic energy is: \[ KE = \frac{1}{2} m v^2 \] Where: - \(KE = 25 \, \text{J}\), - \(m = 0.01 \, \text{kg}\). Substitute values into the equation: \[ 25 = \frac{1}{2} \times 0.01 \times v^2 \] Solving for \(v\): \[ v^2 = \frac{25 \times 2}{0.01} = 5000 \] \[ v = \sqrt{5000} = 25 \, \text{m/s} \] Thus, the velocity of the ball when it hits the ground is 25 m/s.