Respuesta :
Given
Initial velocity:
36 ft/s
Initial height:
0 ft
Vertical motion model:
h(t) = -16t^2 + ut + s
v = initial velocity
s = is the height
Procedure
We are going to use the model provided for the vertical motion.
[tex]\begin{gathered} h(t)=-16t^2+36t+0 \\ h(t)=-16t^2+36t \end{gathered}[/tex]We know that at the maximum height the final velocity is 0.
Then we will use the following expression to calculate the maximum height:
[tex]\begin{gathered} v^2_f=v^2_o-2ah_{\max } \\ 0=v^2_o-2ah_{\max } \\ 2ah_{\max }=v^2_o \\ h_{\max }=\frac{v^2_o}{2a} \\ h_{\max }=\frac{(36ft/s)^2}{2\cdot32ft/s^2} \\ h_{\max }=20.25\text{ ft} \end{gathered}[/tex]Now for time:
[tex]\begin{gathered} 20.25=-16t^2+36t \\ 16t^2-36t+20.25=0 \end{gathered}[/tex]Solving for t,
[tex]\begin{gathered} t_1=2.25 \\ t_2=0 \end{gathered}[/tex]The total time the kangaroo takes in the air is 2.3s.
