Q16DQ - University Physics with Modern Physics

Question

Prove these statements: (a) As long as you can ignore the

effects of the air, if you throw anything vertically upward, it will

have the same speed when it returns to the release point as when it

was released. (b) The time of flight will be twice the time it takes

to get to its highest point.

Step-by-Step Solution

Answer

(a) If air resistance is ignore then $F_{net}=mg.$

Let, velocity of launch = u

       velocity of impact = v

using conservation of energy,

$$\begin{gathered} E_1=E_2 \\ \Rightarrow \frac{m}{u^2}+mg\left(o\right)=\frac{m}{v^2}+mg\left(o\right)v \\ \Rightarrow \frac{m{u}^2}{2}=\frac{m{v}^2}{2}\Rightarrow u=v \end{gathered}$$

Hence, the speed at impact is same as the speed of launch.

 

      (b) Let, time of ascent  = $t_1$

        time of descent = $t_2$

 At maximum height the speed = o

 Acceleration = -g

 

 During moving up:

$$\begin{gathered} v=u+at \\ o=u-g{t}_1 \\ {t}_1=\frac{u}{g} \end{gathered}$$ 

         

 During moving down:

  Initial velocity = o

  Final velocity = -v = -u

$$\begin{gathered} v=u+at \\ -u=o-g{t}_2 \\ {t}_2=\frac{u}{g} \end{gathered}$$        

           

Time of flight = $t_1+t_2=\frac{u}{g}+\frac{u}{g}=\frac{2u}{g}=2\left(\frac{u}{g}\right)$

$\boxed{(t=2t_1)}$             

Hence, time of flight is twice the time.