# Kinematics question

1. Oct 23, 2007

### vericolorful

1. The problem statement, all variables and given/known data
A ball is released while traveling at 15m/s east at an altitude of 10m.
How many meters in front of the target must the ball be released?
What's the velocity of the ball at impact?

2. Relevant equations
d = vt
V$$_{}f$$ = V$$_{}i$$ + at
d = V$$_{}i$$t + .5at$$^{}2$$

3. The attempt at a solution
I know the origin must be 10m above ground level. However, after that I'm not sure what to do. I feel like there's not enough information for me to solve this. This could be a typo on the worksheet. Other problems I did like this gave angles so I'm a little confused without the angle.

2. Oct 23, 2007

### malty

I think although I may be wrong that question is asking that the target is a ground target, and that question is asking how far in front of the target the ball should be dropped from to make it land on the target.

3. Oct 23, 2007

### vericolorful

That's what I'm thinking too, but I don't even know how to approach the question in order to solve it.

4. Oct 24, 2007

### rl.bhat

Initial velocity of the ball towards ground is zero. Using the third equation the time taken by the ball to reach the ground and the velocity with which it hits the ground can be calculated. t = 1.429s and velocity = 14m/s. During this time the ball is moving towards east at 15m/s. Therefore to hit the target, it must be at a distance 1.429x15 = 21.44 m.
The velocity with which it hits the ground is v = (15^2 + 14^2)^1/2 =20.52m/s

5. Oct 24, 2007

### Hootenanny

Staff Emeritus
Your help is much appreciated, but please don't post complete solutions, especially incorrect ones.