Respuesta :
Given data:
* The height of the balcony is 10 m.
* The initial velocity of the from the balcony is 14 m/s.
Solution:
(a). By the kinematics equation, the height of the maximum point is,
[tex]v^2-u^2=-2gh[/tex]Here the negative sign before the g is indicating the motion of ball opposite to the gravitational force.
where v is the final velocity of the ball at the maximum point, u is the initial velocity of the ball from the balcony, g is the acceleration due to gravity, and h is the height of the maximum point from the balcony,
Substituting the known values,
[tex]\begin{gathered} 0-14^2=-2\times9.8\times h \\ 196=19.6\times h \\ h=\frac{196}{19.6} \\ h=10\text{ m} \end{gathered}[/tex]The final velocity of the ball on reaching the ground from the maximum point is,
[tex]v^2_f-v^2_i_{}=2g(h+10)_{}[/tex]where h+10 is the total height of the maximum point from the ground,
v_f is the final velocity on reaching the ground, v_i is the initial velocity of the ball at the maximum point,
As the velocity of the ball at the maximum point becomes zero,
Thus,
[tex]\begin{gathered} v^2_f-0=2\times9.8\times(10+10)_{} \\ v^2_f=392 \\ v_f=19.8\text{ m/s} \end{gathered}[/tex]Thus, the velocity with which the ball hits the ground is 19.8 m/s.
(b). The time taken by the ball from balcony to the maximum point is,
[tex]\begin{gathered} v-u=-gt \\ 0-14=-9.8\times t \\ t=\frac{14}{9.8} \\ t=1.43\text{ s} \end{gathered}[/tex]The time taken by the ball from the maximum point to the ground is,
[tex]\begin{gathered} v_f-v_i=gt^{\prime} \\ 19.8-0=9.8\times t^{\prime} \\ t^{\prime}=\frac{19.8}{9.8} \\ t^{\prime}=2.02\text{ s} \end{gathered}[/tex]Thus, the total time the ball spend in the air is,
[tex]\begin{gathered} T_{}=t+t^{\prime} \\ T=1.43+2.02 \\ T=3.45\text{ s} \end{gathered}[/tex]Hence, the total time spend by the ball in the air is 3.45 s.
