Projectile motion (No initial velocity or time)

1. Sep 10, 2006

PascalPanther

This is the problem:
Let's ignore the second part, because I can't even get to there yet (sounds simple enough if I figured out the first part).

What I am stuck on, is the fact I have no initial velocity or time. So I have two unknowns.
Here is what I think I know. The acceleration is 9.8 m/s^2 due to gravity. Since I have the same y position, at the highest point P, where the velocity is 0, it should be exactly in the middle? So 100m.
If I use, 100m as x when y is at it's highest, I can turn that into a right angle. And do: (100)*tan 45 degrees = 100m. So the height at point P is 100m.

I'm assuming these are the only equations I am suppose to use:
v= v(i) + at
x = x(i) + v(i)t + 1/2at^2
v^2 = v(i)^2 + 2a(x-x(i))

x=100m
y=100m
t=?
v(i)=?
a=9.8 m/s^2

and also for y, since they are independent. And I think x and y have the same answer in this case, because they both have the same magnitude (at least cut in half). However, it doesn't look like I can solve for anything in any of the equations.

What should I be doing next?

2. Sep 10, 2006

E=m(C)^2

Basically for question 1 you need to have separate the information into a vertical component and a horizontal component.
Vertical: u=?, v=0, a=-9.8 and x=100, as you calculated. From here use equation v^2=(u^2)+2as, where u=initial velocity and s=displacement. From this you schould get u=44.3m/s.
Now Horizontal: velocity is always constant horizontally so u=v and a=0. So use the formula s=ut+0.5a(t^2) and substitute values and transpose to get t. You should end up with t=4.51 secs.

3. Sep 10, 2006

tony873004

In your example, wouldn't u=44.3 be the total initial velocity. Then he'd need to extract the x-component for the next step giving a final time in the 6 to 7 second range?

4. Sep 10, 2006

E=m(C)^2

Too true tony, i forgot to breakdown the u=44.3 into a horizontal component. Once broken down it should equal 31.32m/s and this correct value for u leads to t=6.38m/s. Sorry guys