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person throws a ball horizontally from the top of a building that is 24.0 m above the ground level. The ball lands 100 m down range from the base of the building. What was the initial velocity of the ball? Neglect air resistance and use g = 9.81 m/s2.

Respuesta :

Answer:45.24 m/s

Explanation:

Given

Height of Building h=24 m

Range of ball R=100 m

Considering Vertical motion of ball

using [tex]y=u_yt+\frac{a_yt^2}{2} [/tex]

initial vertical velocity is zero therefore [tex]u_y=0[/tex]

[tex]24=0+\frac{9.8\times t^2}{2}[/tex]

[tex]t=\sqrt{\frac{48}{9.8}}[/tex]

[tex]t=2.21 s[/tex]

Now considering Horizontal Motion

[tex]R=u_xt+\frac{a_xt^2}{2}[/tex]

[tex]100=u_x\times 2.21+0[/tex]  , as there is no horizontal acceleration

[tex]u_x=45.24 m/s[/tex]

                       

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