Question

Two rocks are thrown from the top of a very tall tower. One of them is thrown vertically up with an initial velocity of vup = 15.4 m/s. The other rock is thrown horizontally to the right with an initial velocity of vright = 11.5 m/s. (See figure.) How far will the rocks be from each other after 4.08 s? (Neglect air resistance and assume that the rocks will not hit the ground or the tower.)

Answer 1

After 4.08s the vertical rock is s=ut-0.5gt² =15.4x4.08-0.5x9.8x4.08²
=-18.73 m from the starting point (+18.73m below the

starting point). Referred to horizontal and vertical axes,
origin the starting point, its coordinates are (0, -18.73).
The horizontal rock will be 11.50x4.08=46.92 m to the right,
and 0.5x9.8x4.08²=81.56 m downwards. Its cords
are (46.92, -81.56).

(Distance apart)²=46.92²+(-81.56
+18.73)²=6150 and didtance apart =78.42 m

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