1. ## projectile motion help

A ball is dropped from rest from a height h. Neglecting air resistance and taking the acceleration due to gravity to be g, calculate how ling it takes to hit the floor in terms of g and h.

so i set up two equations of motion.

$F = \frac{d}{dt}(mv) = -mgj$

$-mgj = m\frac{d^2x}{dt^{2}} i + m \frac{d^{2}y}{d^{2}}j$

there is not motion in the x direction so

$\frac{d^2x}{dt^{2}} = 0$

$\frac{d^{2}y}{d^{2}} = -g$

so integrating twice i get two equations

$x(t) = x_{1}t + x_{0}$

$y(t) = -\frac{1}{2}gt^{2} + y_{1}t + y_{0}$

not sure where to go from here, any help appreciated, how do i work out the constants on integration ? I know the ball is dropped from rest, so one of the constants will be 0?

2. ## Re: projectile motion help

What does "from rest" mean to you?

It is impossible to answer the question without know what you mean by "y"? If it is the height above the floor then obviously "hitting the floor" means y= 0. Set y equal to 0 and solve for t.

3. ## Re: projectile motion help

You have \frac{dy}{dt}=0 when t=0 ("the ball is dropped from rest") and y=h when t=0 ("from a height h", assuming y=0 is the ground). These two conditions give you y_1 and y_0. You then have to solve y(t)=0 for t, which will be "how long it takes to hit the floor".

- Hollywood

4. ## Re: projectile motion help

Originally Posted by Tweety
so integrating twice i get...

$y(t) = -\frac{1}{2}gt^{2} + y_{1}t + y_{0}$
Please define what you mean by $y_1$. I assume you mean that $y_1$ is the initial velocity of the projectile, which you might show as either $y'(0)$ or $v_0$. In any event it equals 0, and y_0 = h, so you have:

$y(t) = -\frac{1}{2}gt^{2} + h$

Set y(t) = 0 and solve for t.

5. ## Re: projectile motion help

thank you, yes sorry y1 is the initial velocity, and y0 = h .

6. ## Re: projectile motion help

And since you were told that the ball is "dropped from rest", y1= what?