An elevator ascends from the ground with uniform speed. At time T1 a boy drops a marble through a hole in the floor. The marble hits the ground T2 seconds later. Find the height of the elevator at time T1. (Hint: If T1 = T2 = 4 s, h = 39.2 m) How can I solve this? I've worked on it for hours and it's frustrating me.
You have two unknowns, the uniform speed, v0, and the height h. You need two equations, write one for T1 in terms of h and v0 and the other for T2 in terms of h and v0. -Dale
What may cause some confusion here is that T<SUB>1</SUB> is the time as counted from 0, but T<SUB>2</SUB> as counted from T<SUB>1</SUB>. In dependence of a general time variable t, the equation of motion of the marble (after it has been released at time T<SUB>1</SUB> ) is h = v*t -1/2*g*(t-T<SUB>1</SUB>)<SUP><SUP>2</SUP></SUP> where h is the height, v the velocity of the elevator and g the gravitational acceleration. At time t=T<SUB>1</SUB>+T<SUB>2</SUB>, the marble is supposed to hit the ground again, i.e. 0 = v*(T<SUB>1</SUB>+T<SUB>2</SUB>) -1/2*g*T<SUB>2</SUB><SUP><SUP>2</SUP></SUP> , or v*(T<SUB>1</SUB>+T<SUB>2</SUB>) = 1/2*g*T<SUB>2</SUB><SUP><SUP>2</SUP></SUP> , or v*T<SUB>1</SUB>*(1+T<SUB>2</SUB>/T<SUB>1</SUB>) = 1/2*g*T<SUB>2</SUB><SUP><SUP>2</SUP></SUP> . Now since v*T<SUB>1</SUB> = h<SUB>1</SUB> (the height at which the marble is released), you have therefore h<SUB>1</SUB>*(1+T<SUB>2</SUB>/T<SUB>1</SUB>) =1/2*g*T<SUB>2</SUB><SUP><SUP>2</SUP></SUP> , or h<SUB>1</SUB> = g*T<SUB>2</SUB><SUP><SUP>2</SUP></SUP> /2 /(1+T<SUB>2</SUB>/T<SUB>1</SUB>) , which is the result you want. Thomas
h = v*t -1/2*g*(t-T1)2 Thanks Thomas. My only problem then was with the above equation. I suspected that what was needed was to include vt in the equation for the marble's motion. Could you please explain how it is that the marble begins with an initial velocity v? I was under the impression that since it was dropped, the inital velocity would be 0.
The initial velocity is zero -- with respect to the person holding the marble, that is. With respect to the surface, it is v. Think of it this way: Imagine you are riding a high speed elevator and you let marble go as you are riding up. Does the marble instantly acquire a velocity of zero with respect to the surface of the Earth and punch through the elevator floor? No. It drops to the floor of the elevator with more-or-less the same behavior as if the elevator were not moving at all. This is just Newton's First Law of Motion.
As DH mentioned already, in the reference frame of the earth, the marble has the velocity v of the elevator when it is released. So in this frame, the marble is actually not going straight down after its release, but it rises even further until its kinetic energy is used up by gravity, and only then does its height get smaller again. It is very much like throwing the marble into the air from the ground with velocity v. The only difference here is that it is thrown from height h<SUB>1</SUB> (in this case the elevator provides the velocity v). Thomas
