# Projectile Motion

1. Jan 17, 2004

### amyalex

Hi:
This is my first visit to this website. I am a 43-year-old undergraduate majoring in Environmental Horticulture and somehow missed physics along the way. It's been 2-3 years since I had all my algebra, trig and pre-calc and now in my first 2 weeks of physics I find I need help. We just studied projectile motion and the problems that go with it. I think I understand the concepts and can find the answer to a problem that deals with how far the projectile travels given an inital velocity and angle. However, the problem I'm trying to do now the distance is given and I need to come up with the time and the velocity. I actually have my instructor's notes how to solve this, but his notes seem more convoluted than what it necessary based on other websites I have seen. Also, I notice that everyone uses different formulas or at least different symbols in those formulas and it is all very confusing. At any rate, the problem is A daredevil jumps a canyon 15 m wide by driving a motorcycle up an incline sloped at an angle of 37 degrees with the horizontal. What minimum speed must she have in order to clear the canyon? How long will she be in the air? He starts out with the horizontal being Vo = 15/.8t and I'm ok with that until later. Then for the vertical, working from the equation y = Voyt - 1/2 a(or g in this case)t^2, my professor works to Vo sin 37 deg. t - 5t^2 = 0. Here's where I get confused. Does the 5 in the 5t^2 come from 1/2 of 10 m/s for g and how does this equation equal 0 and what is the significance of that? He then goes on to substitutions and other things that I pretty much understand, but I can't get past this zero business.

Thanks for your help and this probably won't be my last post this semester!

Amy
Gainesville, Florida

2. Jan 17, 2004

### himanshu121

Yes 5=10/2

Okay what is the displacement in y direction

3. Jan 17, 2004

### HallsofIvy

Staff Emeritus
A more accurate figure for the "acceleration due to gravity" is 9.8 m/s2 but 10 is okay for an example. I wouldn't want to be that "rough" in planning how to jump a canyon but I guess making the initial speeder higher than necessary is the way to go!

4. Jan 17, 2004

### amyalex

Thanks for answering my question, but each of you has just answered part of it. Can you answer the other part of my question, because that's the part I'm really having trouble with. And by the way, the professor is having us use 10 m/s^2 instead of 9.8.

5. Jan 17, 2004

### NateTG

That equation gives the height of the projectile at time t.

The height is 0 at the beginning and at the end of the flight. To find out how long the motocycle was in the air, you can calculate the time it takes for the equation to be zero.

6. Jan 17, 2004

### himanshu121

The basic formula is
$$y-y_0=v_ot+\frac{at^2}2$$

here y=position of particle after time t
y0= initial position
here a=-g

Now after landing the other side of the cannon the position will be same as initial therefore y=y0

Hence u get y-y0= 0