You throw a ball of mass $\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{}\mathbf{kg}$straight up. You observe that it takes$\mathbf{3}\mathbf{.}\mathbf{1}\mathbf{}\mathbf{s}$to go up and down, returning to your hand. Assuming we can neglect air resistance, the time it takes to go up the top is half the total time, $\mathbf{1}\mathbf{.}\mathbf{55}$ s . Note that at the top the momentum is momentarily zero, as it changes from heading upward to heading downward. (a) Use the momentum principle to determine the speed that the ball had just after it left your hand. (b) Use the Energy Principle to determine the maximum height above your hand reached by the ball.

- The mass of a ball is $1.2\mathrm{kg}$

- The time taken to go up and down is $3.1\mathrm{s}$

- The time takes to go up the top is half the total time, $1.55\mathrm{s}$

- At the top the momentum is momentarily zero

The principle of momentum states that, when the two objects collide, the collision at initial is equal to the collision at the final of the two objects.

The total net force by the principle of momentum,

$F_{\text{net}}=\frac{\Delta P}{\Delta t}\cdots \cdots \left(1\right)$

Where,$\Delta P=$Change in momentum;$\Delta t=$change in time

Momentum is a multiplication of mass and velocity. So the above equation is written as,

$F_{\text{net}}=\frac{m\left(v_f-v_i\right)}{t_f-t_i}\cdots \cdots \left(2\right)$

Substitute $F=mg$in Equation (2),

$mg=\frac{m\left(v_f-v_i\right)}{t_f-t_i}$

where,$v_i,v_f=$initial and final speed

$t_i,t_f=$initial and final time

$g=$acceleration due to gravity $=9.81\mathrm{m}/{\mathrm{s}}^2$

Since the final velocity and initial time is zero. The above equation will become,

$-9·81=\frac{\left(0-V_i\right)}{1.55-0}$

$v_j-9.81\times 1.55$

$=15.21\mathrm{m}/\mathrm{s}$

Hence, the speed of a ball using momentum principle is $15.21\mathrm{m}/\mathrm{s}$

The maximum height by using Energy Principle,

$\frac{1}{2}m{v}_i^2=mgh$

$\frac{1}{2}\times 1·2\times 15·{21}^2=1·2\times 9·81\times h$

Hence, the baseball goes $11.79\mathrm{m}$ height.