horizontal distance (not displacement vector)
hd= sqrt (330^2 + 320^2)
vertical distance
= 10
total distance = sqrt(hd^2+100)
A man leaves his front door, walks 330 m east and 320 m north, stopping at the edge of a cliff. He then takes a penny from his pocket and drops it from the cliff, which is 10 m high. What is the magnitude of the displacement (in meters) of the penny?
I'm confused on how a drawing of this would look and then how would you solve it. Would the vertical leg end up being 310 m and then you find the hypotenuse?
2 answers
the 330 and 320 form a right triangle
the hypotenuse is the horizontal displacement
the horizontal and vertical displacements also form a right triangle
since the three directions are orthogonal
... 330^2 + 320^2 + 10^2 = d^2
the hypotenuse is the horizontal displacement
the horizontal and vertical displacements also form a right triangle
since the three directions are orthogonal
... 330^2 + 320^2 + 10^2 = d^2
1 answer