Question

An object falls from rest. In the last second of its fall it covers half of the total distance. If g=9.8m/s^-2 then the total time of its fall is
A)(2-2^1/2)s
B)(2+2^1/2)s
C)(2+-2^1/2)s
D)4s

Answer 1

Using a SUVAT eqn, displacement \(s(t) = ut + \frac{1}{2} a t^2\), BUT with \(u = 0\) as object falls from rest. So: \(s(t) = \frac{1}{2} a t^2\).

As \(a = g\), with downward velocity positive, if the total travel time is \(T\) then:

\(\qquad s(T) = \frac{1}{2} g T^2\)

Equally, distance travelled up to "start of last second" is:

\(\qquad s(T-1) = \frac{1}{2} g (T-1)^2\)

So, distance travelled in last second is:

\(\qquad s(T) - s(T-1) = \frac{1}{2} g \left( T^2- (T-1)^2 \right)\)

\(\qquad = \frac{1}{2} g \left( 2T - 1 \right)\)

Using given ratio:

\(\qquad \dfrac{\frac{1}{2} g \left( 2T - 1 \right)}{ \frac{1}{2} g T^2} = \dfrac{1}{2}\)

\(\qquad \implies T^2 - 4T + 2 = 0\)

Leaving you to solve that (using the quadratic formula) to get solution C)

Answer 2

Thank you!

Answer 3

An interesting way to do this problem is to use the conservation of mechanical energy

Answer 4

Oh, okay
Thank you!