Question

Time taken by the ball to reach the ground.
Velocity of the ball on reaching the ground.

Hint:

For objects in free fall starting from rest, the equations of motion simplify. Remember that the mass of the object does not affect the time of fall or the final velocity in the absence of air resistance.

Answer 1

Given data: Mass (m) = 3 kg, Height (s) = 125 m, Initial velocity (u) = 0 m/s (since it's released), Acceleration due to gravity (g) = 10 m/s$^2$. Time taken by the ball to reach the ground (t): We use the second equation of motion: \[ s = ut + \frac{1}{2}gt^2 \] Substituting the given values: \[ 125 = (0 \times t) + \frac{1}{2} \times 10 \times t^2 \] \[ 125 = 5t^2 \] \[ t^2 = \frac{125}{5} = 25 \] \[ t = \sqrt{25} = 5 \, \text{s} \] So, the time taken by the ball to reach the ground is 5 seconds. Velocity of the ball on reaching the ground (v): We use the first equation of motion: \[ v = u + gt \] Substituting the values: \[ v = 0 + (10 \times 5) \] \[ v = 50 \, \text{m/s} \] Alternatively, using the third equation of motion: \[ v^2 = u^2 + 2gs \] \[ v^2 = 0^2 + 2 \times 10 \times 125 \] \[ v^2 = 2500 \] \[ v = \sqrt{2500} = 50 \, \text{m/s} \] So, the velocity of the ball on reaching the ground is 50 m/s.