1) Maximum height
This is a uniformly accelerated motion with initial speed [tex]v_0 = 8.8 m/s[/tex] and acceleration [tex]a=g=-9.81 m/s^2[/tex] (the negative sign is required because the acceleration is directed downward, in the opposite direction of the motion, which is directed upward). The distance covered (S) in a uniformly accelerated motion is related to the acceleration and the initial and final velocity by the relationship
[tex]2aS=v_f^2-v_i^2[/tex]
At maximum height, S=h and the final velocity is zero: [tex]v_f =0[/tex]. So we can solve for h and find the maximum height:
[tex]2ah=-v_i^2[/tex]
[tex]h= \frac{-v_i^2}{g}= \frac{-(8.8 m/s)^2}{-9.81 m/s^2}=3.95 m [/tex]
2) Speed at half maximum height
Half maximum height corresponds to [tex] \frac{h}{2}= \frac{3.95 m}{2}=1.975 m [/tex]. To find the speed at this height, we can use the same relationship used in the previous step:
[tex]2a \frac{h}{2}=v_f^2-v_i^2 [/tex]
where [tex]v_f [/tex] is the speed at half maximum height.
Re-arranging and substituting numbers, we find
[tex]v_f = \sqrt{v_i^2+(2g \frac{h}{2}) }= \sqrt{(8.8m/s)^2+2(-9.81m/s^2)(1.975m)}=6.22m/s [/tex]
