# Homework Help: Velocity equation help

1. Oct 12, 2007

### klm

You're 6.0 m from one wall of a house. You want to toss a ball to your friend who is 6.0 m from the opposite wall. The throw and catch each occur 1.0 m above the ground. V Assume the overhang of the roof is negligible, so that you may assume the edge of the roof is 6.0 m from you and 6.0 m from your friend.

What minimum speed will allow the ball to clear the roof?

okay i figured out this much so far:
delta x= 9m
9= (Vo cos theta)T
delta y=5
5= (Vo sin theta)T- .5gt^2

Vy=Vosintheta - gt=0

t=Vosintheta / g

and now i know i am suppose to plug in T into the x equation and y equation to solve for Vo, but i cant figure out to do that?

2. Oct 12, 2007

### klm

like i know if you substitute t in the first equation you can get,

9= (Vo cos theta) (Vosintheta / g) so does that equal
9= (2Vo costheta sin theta) / g

and then

5= (Vo sin theta) (Vosintheta / g) - .5g (Vosintheta / g)^2
5= Vosin2theta/2g

but then what?

3. Oct 12, 2007

### klm

4. Oct 13, 2007

### klm

if anyone could please look at my work that would be great. i am online to discuss. please.

5. Oct 13, 2007

### klm

thank you first of all for responding! but i am very confused as what i am suppose to do? i know this much, that t=Vosintheta / g, but where am i suppose to substitute that into?

6. Oct 13, 2007

### FedEx

See here you know the maximum height and that is 3 + 3tan(45).

Now the range is 18 m.

Hence we have two variables and two equations.

So simultaneously solve the following two.

$$H_{max}= \frac{v^2sin^2\theta}{g}$$

$$R = \frac{v^2\sin2\theta}{g}$$