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, $m = 400 \text{ gm} = 0.4 \text{ kg}$ Height of the tower, $h = 90 \text{ m}$ Initial velocity, $u = 0 \text{ m/s}$ (since it is dropped) Acceleration due to gravity, $g = 9.8 \text{ m/s}^2$ i) Time to reach the ground ($t$): Step 1: Use the kinematic equation relating displacement, initial velocity, time, and acceleration. $$ h = ut + \frac{1}{2}gt^2 $$ Step 2: Substitute the given values. $$ 90 \text{ m} = (0 \text{ m/s})t + \frac{1}{2}(9.8 \text{ m/s}^2)t^2 $$ $$ 90 = 4.9t^2 $$ Step 3: Solve for $t$. $$ t^2 = \frac{90}{4.9} $$ $$ t^2 \approx 18.3673 \text{ s}^2 $$ $$ t = \sqrt{18.3673} $$ $$ t \approx 4.286 \text{ s} $$ The time to reach the ground is $\boxed{\text{4.29 s}}$. 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. $$ v^2 = (0 \text{ m/s})^2 + 2(9.8 \text{ m/s}^2)(90 \text{ m}) $$ $$ v^2 = 0 + 1764 \text{ m}^2/\text{s}^2 $$ $$ v = \sqrt{1764} $$ $$ v = 42 \text{ m/s} $$ The velocity with which it strikes the ground is $\boxed{\text{42 m/s}}$. Step 3: Calculate the final momentum ($p$). $$ p = mv $$ Step 4: Substitute the mass and final velocity. $$ p = (0.4 \text{ kg})(42 \text{ m/s}) $$ $$ p = 16.8 \text{ kg m/s} $$ The final momentum is $\boxed{\text{16.8 kg m/s}}$. That's 2 down. 3 left today — send the next one.