Question #125787

A ball is thrown horizontally at a velocity of 10.0 m/s from the top of a 90.0 m building. Calculate the distance from the base of the building that the ball is safely caught on the ground.

Expert's answer

Horizontal velocity will be 10 m/s, initial vertical velocity will be 0.

Now, For horizontal motion, at any time t,

"x = vt"

For vertical motion,

"y = \\frac{1}{2}gt^2"

From both equations, removing t

we get,

"y = \\frac{1}{2}g (\\frac{x}{v})^2 = \\frac{g}{2v^2}x^2"

"\\implies x = \\sqrt{\\frac {2 v^2 y}{g}} = (\\sqrt{\\frac {2 y}{g}})v"

Now putting the values,

"x = (\\sqrt{\\frac {2 *90}{10}})10 = 42.43 m"

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