A ball of mass 400 gm is dropped from tower of height 90 m. Find: i) Time to reach the ground. ii) Velocity with which it strike the ground and its final momentum.

Here's the solution for question 13: 13. A ball of mass 400 gm is dropped from tower of height 90 m. Find: i) Time to reach the ground. ii) Velocity with which it strike the ground and its final momentum. Given: Mass of the ball, m = 400 gm = 0.4 kg Height of the tower, h = 90 m Initial velocity, u = 0 m/s (since it is dropped) Acceleration due to gravity, g = 9.8 m/s^2 i) Time to reach the ground (t): Step 1: Use the kinematic equation relating displacement, initial velocity, time, and acceleration. h = ut + (1)/(2)gt^2 Step 2: Substitute the given values. 90 m = (0 m/s)t + (1)/(2)(9.8 m/s^2)t^2 90 = 4.9t^2 Step 3: Solve for t. t^2 = (90)/(4.9) t^2 ≈ 18.3673 s^2 t = sqrt(18.3673) t ≈ 4.286 s The time to reach the ground is 4.29 s. ii) Velocity with which it strikes the ground (v) and its final momentum (p): Step 1: Calculate the final velocity (v) using the kinematic equation. v^2 = u^2 + 2gh Step 2: Substitute the given values. v^2 = (0 m/s)^2 + 2(9.8 m/s^2)(90 m) v^2 = 0 + 1764 m^2/s^2 v = sqrt(1764) v = 42 m/s The velocity with which it strikes the ground is 42 m/s. Step 3: Calculate the final momentum (p). p = mv Step 4: Substitute the mass and final velocity. p = (0.4 kg)(42 m/s) p = 16.8 kg m/s The final momentum is 16.8 kg m/s. That's 2 down. 3 left today — send the next one.