Apply the general results obtained in the full analysis of motion under the influence of a constant force in Section 2.5 to answer the following questions. You hold a small metal ball of mass m a height h above the floor. You let go, and the ball falls to the floor. Choose the origin of the coordinate system to be on the floor where the ball hits, with y up as usual. Just after release, what are y_i and v_yi? Just before hitting the floor, what is y_f? How much time ∆t does it take for the ball to fall? What is v_fy just before hitting the floor? Express all results in terms of m, g, and h. How would your results change if the ball had twice the mass?

• The initial location of the ball is given by, y_i
• The final location of the ball is given by, y_f
• The initial speed of the ball is given by, V_yi
• The final speed of the ball is given by, V_yf

The time taken by ball to fall is given by,$∆t$

The following is an explanation of conservation of momentum. There is a perfect match between total momentum of two particles before and after a collision occurs inside an isolated system between particle 1 and particle 2.

The principle of momentum is given by,

$V_{yf}=V_{yi}+\frac{F_{net.y}}{m}∆t...........\left(1\right)$

The force acting on the ball is purely gravitational force, here we assume that the speed of the ball is negligible, the net force on the ball can be given by,

$F_{net.y}=F_g=-mg$

Here, the initial location of the ball is $y_i=h$

The initial speed of the ball is$V_{yi}=0$

The final location of ball before it hits the ground is $y_f=0$

The time needed by ball to fall can be evaluated position equation,

role="math" localid="1654005095363" $y_f=y_i+V_{yi}∆t+\frac{1}{2}\frac{F_{net.y}}{m}∆t^2$

Substitute the values in equation (1),

$0=h+0+\frac{1}{2}\frac{F_{net.y}}{m}∆t^2$

$∆t=\sqrt{\frac{2h}{g}}$

Thus, the time taken by the ball to fall is $\sqrt{\frac{2h}{g}}$.

The final velocity of the ball can be evaluated using equation (1),

$$\begin{gathered} V_{yf}=0-g\sqrt{\frac{2h}{g}} \\ {V}_{yf}=-\sqrt{2gh} \end{gathered}$$

If the ball has twice the mass the result won’t change as it does not depend on the mass.

Thus, there is no change in result of the ball.