the cable holding an empty elevator breaks when the elevator is at rest at the highest point of a 120 m building. With what velocity (magnitude and direction) will the elevator hit the ground (terminal velocity)
Answers
Answer:
We choose down as the +y direction
and use the equations of Table 2-1 (replacing X with y) with
a = +g, V0 =0, and y0 =0,
We use subscript 2 for the elevator reaching the ground and 1 for the halfway point.
(a) with 3rd equation of motion
V22 = V02+2a(y2−y0),
putting values
V2 = 2gy2 = 2(9.8m/s2)(120m) = 48.5 m/s.
(b) The time at which it strikes the ground is
using 2nd equation of motion
t2 = g2y2
= 9.8m/s22(120m) = 4.95 s.
(c) with 3rd equation of motion
V12=V02+2a(y1−y0),
putting values
V1 = 2gy1
Answer:
Velocity will come by THIRD Equation of Motion.
v²= u²+2aS
v²= 2×9.8×120
v= 48.4974
Direction depends upon frame of reference..but if you taking along gravity then + (towards bottom) or if against gravity then -(towards bottom).
48.4974 is the terminal velocity.