Answer:
a) t = 0.782
b) v_b/el = -7.67 m/s
c) v_b/e = - 5.17 m/s
d) y_b/e = 1.04 m down-wads
Explanation:
Given:
- initial position of bolt y_b,i = 3.0 m
- initial velocity of bolt v_b,i = 2.50 m/s
- constant velocity of elevator v_e = 2.50 m/s
- acceleration of free fall for bolt a = 9.81 m/s^2
Find
- (a) How long does it take for the bolt to fall to the elevator floor? What is the speed of the bolt just as it hits the elevator floor
- (b) according to an observer in the elevator?
- (c) According to an observer standing on one of the floor landings of the building?
- (d) According to the observer in part (c), what distance did the bolt travel between the ceiling and the floor of the elevator?
Solution:
- The position of bolt y_b is given by kinematic equation of motion:
y_b = 3.0 + 2.5*t +0.5*9.81*t^2
- position of floor with constant upward speed is:
y_f = 2.50*t
- When bolt hits the floor they have same position:
3.0 + 2.5*t +0.5*9.81*t^2 = 2.50*t
4.905*t^2 = 3.0
t = sqrt(3/4.905) = 0.782 s
- velocity of the bolt relative to earth is:
v_b/e = 2.50 - 9.81*0.782
v_b/e = -5.17 m/s
- Hence, relative to observer in elevator:
v_b/el = -5.17 - 2.5
v_b/el = -7.67 m/s
- Relative to earth the distance traveled by bolt is:
y_b/e = 2.5 - 0.5*9.81*(0.782)^2
y_b/e = -1.04 m (downwards)
