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KATRIN_1 [288]
3 years ago
9

1: A baseball is hit 4 feet above the ground leaves the bat with an initial speed of 98 ft/sec at an angle of 0 45 is caught by

an outfielder at a height of 3 feet.
Engineering
1 answer:
zvonat [6]3 years ago
5 0

Answer:

299.36 feet

Explanation:

To \ find  \   the  \ distance \  of  \ the  \ ball \  from  \ the \ home  \ plate.  \\ \\ From  \ the  \ given  \ information:

Height \ h = 4 \ ft

Initial \ speed \ V_o = 98 \ ft/s ec

The  \ angle \  \theta = 45^0

Acceleration \ due \ to \ gravity (g)= 32.2 \ ft/s

U_x = V_o \ cos 45 = \dfrac{98}{\sqrt{2}}

U_y = V_o \ sin 45 = \dfrac{98}{\sqrt{2}}

So;

S_y = u_y t - \dfrac{1}{2}gt^2

-1 =\dfrac{98}{\sqrt{2}}t - \dfrac{1}{2}*32*1.85t^2

By solving:

t_1 = 4.32 \ sec

Thus;

horizontal \ distance = U_x t

= \dfrac{98}{\sqrt{2}}\times 4.32

\mathbf{=299.36  \ feet}

\mathbf{Thus \ , the  \  distance \ from \  the  \ home  \ plate \  =  \ 299.36  \ feet}

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