Answer:
299.36 feet
Explanation:
[tex]To \ find \ the \ distance \ of \ the \ ball \ from \ the \ home \ plate. \\ \\ From \ the \ given \ information:[/tex]
[tex]Height \ h = 4 \ ft[/tex]
[tex]Initial \ speed \ V_o = 98 \ ft/s ec[/tex]
[tex]The \ angle \ \theta = 45^0[/tex]
[tex]Acceleration \ due \ to \ gravity (g)= 32.2 \ ft/s[/tex]
[tex]U_x = V_o \ cos 45 = \dfrac{98}{\sqrt{2}}[/tex]
[tex]U_y = V_o \ sin 45 = \dfrac{98}{\sqrt{2}}[/tex]
So;
[tex]S_y = u_y t - \dfrac{1}{2}gt^2[/tex]
[tex]-1 =\dfrac{98}{\sqrt{2}}t - \dfrac{1}{2}*32*1.85t^2[/tex]
By solving:
[tex]t_1 = 4.32 \ sec[/tex]
Thus;
[tex]horizontal \ distance = U_x t[/tex]
[tex]= \dfrac{98}{\sqrt{2}}\times 4.32[/tex]
[tex]\mathbf{=299.36 \ feet}[/tex]
[tex]\mathbf{Thus \ , the \ distance \ from \ the \ home \ plate \ = \ 299.36 \ feet}[/tex]
