Question

A test tube is knocked off a tower at the top of a building that is 400 feet above the ground. (For our purposes, we will assume that air resistance is negligible.) The test tube drops 16𝑡^2 feet in 𝑡 seconds.

a) Calculate the average velocity in the last two seconds of the fall.

b) Calculate the instantaneous velocity when the test tube lands.

Answer 1

so first is to find time of fall:
s=400-16t^2
at final time, s is zero
or Tf=sqrt(400/16)=5 seconds.
velocity= ds/dt=-32*t
veloicty(5)=-160 m/s
velocity(3)=-96 m/s
so what is the average velocity?

Answer 2

Thank you, for some reason I thought that 16t^2 was already acceleration (meaning f''(t) ) so I was like "how do I back-track to get velocity" ( f'(t) ).

Anyway, I worked out -128ft/s for V_avg (since the equation is asking for feet/second).
Thus the time hitting the ground was f'(5).

Thanks for the help Bob!

Answer 3

I haven't looked through the calculations, but f'(5) is not time -- it is velocity at t=5.