Help with vector kinematics question

[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?

 

[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.

 

[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.

 

[hisoko]

Yes I tried it too, i tried so many times, it doesn't 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 doesn't work out. z_z lol

 

[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!

 

[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.

 

[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.