# Help With PROJECTILE MOTION

1. Sep 23, 2006

### the_d

***Help With PROJECTILE MOTION***

im giving a problem where i have to find the angle a projectile situated at the top of a will hit an boat wit constant velocity 20 ft/s 5 sec after starting at Vo = 96.6 ft/s and also how high the hill is above the water. The distance from the base of the hill to the boat is given to be 301 ft.

i found the distance the boat travels after 5 seconds to be 100 ft. now im trying to calculate the Vx and Vy components of the projectile to get theta, any suggestions?? thanx

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2. Sep 24, 2006

### FunkyDwarf

Draw a vector triangle and use cos and sine functions. youll see it.

3. Sep 24, 2006

### the_d

i found the angle but how do i find the height of the hill???

4. Sep 24, 2006

Again, which are the equations of displacement? Use the height of the hill as an initial, let's say, y-position.

5. Sep 24, 2006

### the_d

Can someone help me figure out the height of the hill for this problem, i am stuck...thanks

6. Sep 24, 2006

### Staff: Mentor

The projectile follows a parabolic trajectory, but it is starting at a higher elevation than it will end.

One simply finds the point at which the boat and projectile paths intersect at 5 seconds.

The boat travels 100 ft in 5 seconds, the project must travel (310 ft + 100 ft = 401 ft) in those same 5 seconds. Knowing the distance traveled and the time, what value can be calculated?

7. Sep 24, 2006

### the_d

i figured that part out already but now i am having a hard time computing the height of the hill, I tried making a triangle out of the picture but i only have one side 301 and one angle 90 degrees, am i missing somrthing????

8. Sep 24, 2006

### Staff: Mentor

Well, if one has the equation of the parabolic trajectory, then one can determine the time from start to the peak of the trajectory, and from peak of trajectory to impact point. One can then subract the distance traveled from beginning to peak from the peak to impact and that gives the starting elevation.

So one has calculated the angle at the beginning?

Bear in mind, assuming no wind resistance, the horizontal velocity is constant.

9. Sep 24, 2006

Seems it would be a great idea to post a detailed tutorial on projectile motion, since it's such a frequent question.

10. Sep 24, 2006

### tealish

What a smashing idea! Just marvellous

11. Sep 24, 2006

### Staff: Mentor

We'll see what we can do.

Meanwhile, Greg Bernhardt posted a tutorial on "Motion in 2 Dimensions"
Page 11-15 (including worked example) covers projectile motion.

4. Problem-solving strategy for 2–D motion:

a) Select a coordinate system (typically Cartesian).

b) Resolve the initial velocity vector into x and y components.

c) Treat the horizontal motion and the vertical motion independently!

d) Horizontal motion has constant velocity. (Assuming wind resistance is neglected.

e) Vertical motion has constant acceleration (usually gravity).

Afterall, the parabolic trajectory can be described by pair of parametric equations x = x(t) and y = y(t), which describe the location or coordinates as functions of time.
http://archives.math.utk.edu/visual.calculus/0/parametric.6/index.html
http://colalg.math.csusb.edu/~devel/precalcdemo/param/src/param.html
http://www.math.hmc.edu/calculus/tutorials/parametric_eq/