Answer:
a.1.14t
b. 12.084m
c.1.6m
Explanation:
a. Given the vertical information as [tex]y=0m, y_{peak}=?, v_{iy}=5.6m/s, v_{fy}=-5.6m/s, a_y=-9.8m/s ,v_{ypeak}=0 m/s[/tex]
-We use the [tex]v_{fy}=v_{iy}+a_y t[/tex] to determine the time flight as:
[tex]v_{fy}=v_{iy}+a_y t\\-5.6m/s=5.6m/s-9.8m/s^2 t\\\\11.2m/s=9.8m/s^2t\\\\t=1.14s[/tex]
Hence the time of flight is 1.14t
b. Now given the vertical information as [tex]v_{ix}=10.6m/s, \ a_x=0, \ x=?[/tex] we use the equation [tex]x=v_{ix}t+0.5a_xt^2[/tex] to solve for x.
-Note that ax is 0 m/s/s so the last term on the right side of the equation cancels. By substituting 10.6 m/s for vix and 1.1497 s for t , the x can be found to be:
[tex]x=v_{ix} t+0.5a_xt^2\\\\x=10.6m/s\times 1.14+0.5\times 0.0(1.14)^2\\\\x=12.084m[/tex]
Hence the horizontal distance is 12.084m
c. Finally, we use [tex]t_{up}=0.5\times 1.14=0.57s[/tex] and the equation [tex]y_{peak}=v_{iy} t_{up}+0.5a_yt_u_p^2[/tex] with the vertical information [tex]v_{ix}=10.6m/s, \ a_x=0, \ x=?[/tex] to find the y at the peak.
-Substituting and solving we get:
[tex]y_{peak}=v_{iy} t_{up}+0.5a_yt_u_p^2\\\\y=5.6\times0.57+0.5\times -9.8\times 0.57^2\\\\y=1.6m[/tex]
Hence the peak height of the jumper is 1.6m
