Show that the velocity of the particle is... (Newtons Law question)

Question

A particle of mass m is released from rest from height of ten metres above a body of viscous fluid. Show that the velocity of the particle at the moment of impact with the fluid is $14\text{m s}^{-1}$ and estimate the duration of the drop. (Use the approximation $g=10\text{m s}^{-2}$).

So far I have worked out the initial conditions must be $x(0)=10, \dot{x}(0)=0$.

With this my motion for all time is $x(t)=-5t^2+10$

where do i go from here?

Answer 1

Find the time it takes to fall $10$ meters (when $x(t)=0$) by solving:

$$x(t)=0=-5t^2+10$$ to get $t^2=2$. Thus, it takes $\sqrt{2}$ seconds to fall the $10$ meters. Now plug this into the expression for velocity, $\dot x = -10 \sqrt{2}$ to get $14 \frac{m}{s}$ at impact.

Hope this helps,

Paul Safier

Answer 2

Hints/Solution Outline:
Set $x(t) = 0$ to find the time at which it hits the viscous fluid. (This should be $\sqrt{2} \text{s}$.) This is also called the duration of the drop.

Next, find the velocity equation $\dot{x}(t)$. Now, plug in the time that you found from before, and you should have your answer ($\dot{x}(\sqrt{2})$).