Answer:
a) [tex]v_0=640[/tex] ft/s
b) [tex]y=H+1600[/tex] ft where[tex]H[/tex] represents the height as the projectile was launched.
c)[tex]x=11085.13[/tex] ft
Explanation:
First, recognize the values that are given in the problem:
[tex]\alpha =30 ^o[/tex]
[tex]t_f=20[/tex]
[tex]y_f=H[/tex]
a) With those three use this formula: [tex]y=H+v_0sin(\alpha)t-\frac{1}{2}gt^2[/tex]
to find the initial velocity [tex]v_0[/tex].
[tex]\\y_f= H+v_0sin(\alpha)t_f-\frac{1}{2}gt_f^2 \\H= H+v_0sin(30)(20)-\frac{1}{2}(32)(20)^2 \\ -v_0sin(30)(20)=-\frac{1}{2}(32)(20)^2\\ v_0=\frac{\frac{1}{2}(32)(20)^2}{sin(30)(20)} \\ v_0=\frac{6400}{10} =640[/tex]
b) In order to find the maximum altitude, the time is needed to apply the formula. The maximum altitude is when the velocity in the y-axis is equal to zero, so use the formula for the velocity in the y-axis is to find the time, the formula is:
[tex]v_y=v_{0y}-gt\\v_y=v_0sin(\alpha)-gt\\0=(640)sin(30)-32t\\32t=(640)sin(30)\\t=\frac{(640)sin(30)}{32}=\frac{320}{32}=10[/tex]
With [tex]s=10[/tex]s use this formula for the altitude: [tex]y=H+v_0sin(\alpha)t-\frac{1}{2}gt^2[/tex]
[tex]y=H+v_0sin(\alpha)t-\frac{1}{2}gt^2\\y=H+(640)sin(30)(10)-\frac{1}{2}(32)(10)^2\\y=H+3200-1600\\y=H+1600[/tex]
Finally, the range is the maximum displacement in the x-axis, the formula of the displacement is:
[tex]x=v_{0x}t=v_0cos(\alpha )t[/tex]
And the maximum occurs when[tex]t=20[/tex]s
[tex]x=v_0cos(\alpha)t\\x=640cos(30)(20)\\x=11085.13[/tex]
