Distance travelled by a thrown object

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Homework Help Overview

The problem involves a scenario where a snowball is thrown horizontally from a height of 20m at a speed of 25 m/s. Participants are tasked with determining where the snowball lands relative to Point B, which is 50m away horizontally.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the need to find the time of flight before calculating the horizontal distance traveled. There is confusion regarding the time equation mentioned in the original post and its derivation from known equations of motion.

Discussion Status

Some participants are exploring the relationship between the time of flight and the height from which the snowball is thrown. Guidance has been offered regarding the use of gravitational equations to find the time, but there is still uncertainty about the specific equation referenced in the original post.

Contextual Notes

Participants are working under the assumption that the snowball is thrown horizontally and are questioning the validity of the equations provided in the context of the problem. There is a lack of consensus on the derivation of the time equation used in the solution.

Dberry
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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...
 
Physics news on Phys.org
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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