Help with vector kinematics question

1. Oct 2, 2009

hisoko

http://i58.photobucket.com/albums/g280/hisokeee/aa.png

so since the ball is hit 1m above the ground, i will just ignore that and the 21m height fence and just make it a projectile of a ball from ground to the top of a 20m fence

I only know the gravity, the height of fence and distance to fence and the angle the ball was hit. I tried first by figuring out the initial Voy by doing Vfy=0 g=-9.8 and H=20 but that didn't work out. What else can I do with this information?

2. Oct 3, 2009

CompuChip

In this type of problems, you always have two independent directions, namely horizontal and vertical.

First decompose the velocity v into a vertical and horizontal component, using the sine and cosine of the angle.
For the vertical direction, you can then relate v to the final time using the data you gave in your post (cf. $\Delta v = - g t$).
For the horizontal direction, you have information about the distance, which also gives a relation between v and t (cf. $\delta s = v t$).

This will give you two equations involving v and the time t it takes the ball to reach the fence. You can use one of them to eliminate t and obtain a single equation for v.

3. Oct 3, 2009

dave_baksh

I tried this for funsies and can't get the answer. Any chance of further help? Not sure if the OP got it figured out so I don't want an answer, just a little more guidance.

4. Oct 3, 2009

hisoko

Yes I tried it too, i tried so many times, it doesnt get the answer. and with only 2 known things, there really is only one way to go with it because you will have to do the vertical part first using gravity and that answer doesnt work out. z_z lol

5. Oct 3, 2009

Delphi51

It seems to work for me. From the d=vt for the horizontal part, I get an equation with two unknowns, v and t, where v is the initial speed.
From the distance formula for the vertical part, I get another equation with the same two unknowns. Two equations, two unknowns, no problem! Have another go and show your work here so we can help you!

6. Oct 4, 2009

dave_baksh

Success! I get the right answers. That'll teach me to set my working out neatly in future to avoid stupid mistakes! Cheers for the pointers dudes.

Last edited: Oct 4, 2009
7. Oct 4, 2009

CompuChip

Dave, did you pay attention that:
* d in t = d / v is the horizontal distance
* v in t = d / v is the horizontal velocity, and u in s = u t + (1/2) a t^2 is the vertical velocity. Both can be expressed in terms of the speed V using the (co)sine of an angle.