Distance travelled by a thrown object

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AI Thread Summary
A person at Point A throws a snowball horizontally at 25 m/s, and the problem asks where it lands relative to Point B, with the answer being at Point B. The relevant equations for solving this problem include t=(2h/g)^(1/2) for time and d=vt for distance. The confusion arises regarding the time equation, which is derived from the kinematic equation for free fall, s=1/2 gt^2, where s is the height. The discussion highlights the need to determine the time of flight first before calculating the horizontal distance traveled. Understanding the derivation of the time equation is crucial for solving the problem effectively.
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Homework Statement



The diagram to this problem is of a HILL as shown:

Point A
-
-
20m
-
-
------------------------50m------------------------- Point B

From ground level, a person at Point A throws a snowball horizontally to the right at 25 m/s. Where does the snowball land?

a) 10m to the left of Point B
b) At Point B
c) 12.5m to the right of Point B
d) 50m to the right of Point B

Homework Equations



According to the answer given (answer is B), the relevant equations are t=(2h/g)^(1/2) and d=vt

The Attempt at a Solution



The 4 linear motion equations I know are

x = x + vt + (1/2)at^2
v(initial) = v(final) + at
v(i)^2 = v(f)^2 + 2ax
v(avg) = (v(i) + v)/2

I don't know the time equation listed in the answer. Is it derived from one of these equations or is it a whole different equation? I'm confused about how I was supposed to know to solve this question basically...
 
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if you dropped the ball how long does it take?

then use that time to compute how far it went.
 
I know I have to find time first, but if you read the comments I'm asking about the equation used. I have never seen it before and I'm confused about where it comes from. Is it derived from another equation?
 
s= 1/2 g t^2 with s being the height of point A
 
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