Solve Kinematics Question for Ball Released at 15m/s and 10m Altitude

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A ball released at 15 m/s east from a height of 10 m must be dropped 21.44 meters in front of the target to hit it accurately. The time taken for the ball to reach the ground is approximately 1.429 seconds, during which it maintains its eastward velocity. The impact velocity of the ball is calculated to be about 20.52 m/s, combining its horizontal and vertical components. The discussion highlights confusion regarding the lack of an angle in the problem, suggesting it may be a typo. Clarification on the approach to solving such kinematics problems is sought, emphasizing the need for accurate information.
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Homework Statement


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?

Homework Equations


d = vt
V_{}f = V_{}i + at
d = V_{}it + .5at^{}2

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.
Thanks in advance.
 
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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.

Hazy question, I must admit.
 
That's what I'm thinking too, but I don't even know how to approach the question in order to solve it.
 
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
 
rl.bhat said:
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
Your help is much appreciated, but please don't post complete solutions, especially incorrect ones.
 
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