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 = 10.1 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

This should be in physics..... Assuming that gravity exists in this case. g = 9.8 m/s^2


so lets track how high the first rock went.

y=15.6t-0.5g(t)^2

t= 4.08s

so the rock went up 63.648-81.56736 = -17.9136 m up = 17.9 m down

The rock thrown horizontally travelled

x = 10.1t = 41.208m

gravity acting on the rock thrown horizontally pushes it down

so y = 0.5g(t)^2 = 81.56736m

so now we can subtract the two y directions to get the final distance between the two rocks in the y component.

= 81.56736 -17.9136 = 63.5

in the x direction the distance is 41.2 m

so doing the Pythagorean theorem we do

D=sqrt(y^2 + x^2)

so then d = sqrt((41.2)^2 + (63.5)^2)= ?

that should be the final distance between the two rocks.

*Notice that since gravity is the same for both rocks we could have just ignored it and came to the same results.