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 -15.6 m/s. The other rock is thrown horizontally to the right with an initial velocity of Vright -10.3 m/s. (See figure.) V. up V. right How far will the rocks be from each other after 4.11 s? (Neglect air resistance and assume that the rocks will not hit the ground or the tower.) Subt Ane Tries 0/12

Answer 1

For vertical motion Relative initial velocity u = 15 . 6 m/s Relative acceleration a = g - g = 0 m/s^2 therefore After time t = 4.11 s Relative vertical displacement y = 4 times t = 15 . 6 times 4.11 = 64 . 12 m For horizontal displacement therefore x = 10. 3 m/s times 4.11s = 42. 3 m therefore distance between rates d = squareroot (64 . 12)^2 + (42 . 3)^2 d = 76 . 8 m