Question

Box going up elevator uniformly at 5m/'s falls off and hits bottom in 1.6secs

A) what time to reach max height?
B) height at which it fell?
C) height at 0.2 secs after falling

Answer 1

b) First we find the height from which the box fell. Use the formula

$$h\left( t \right)=\frac{g{{t}^{2}}}{2}$$

where g=9.8 m/s² is the acceleration of gravity, t=1.6 secs.Substituting these values, we get

$$h=\frac{9.8\cdot {{1.6}^{2}}}{2}=12.5\,\text{m}$$

So the height at which the box fell is h=12.5 meters.

a) Now we find the time for which the box reached its maximum height. As the box going up elevator uniformly at v=5 m/s the height to which it will rise during time t is

$$h=v\cdot t$$

Substitute h=12.5 m and v=5 m/s

$$12.5=5\cdot t$$

Solving this for t we get the time for which the box reached its maximum height

$$t=\frac{12.5}{5}=2.5\,s$$

c) To find height at 0.2 secs after falling first we find the distance that the box will cover in 0.2 seconds

$$s\left( 0.2 \right)=\frac{g{{t}^{2}}}{2}=\frac{9.8\cdot {{0.2}^{2}}}{2}=0.2\,\text{m}$$

and then subtract this distance from the height of the fall. We get

$$h\left( 0.2 \right)=12.5-0.2=12.3\,\text{m}$$

So height at 0.2 secs after falling is 12.3 m