Projectile Motion: How tall is the building?

Click For Summary

Homework Help Overview

The problem involves a ball thrown horizontally from the roof of a building, with the goal of determining the height of the building based on the horizontal distance traveled before landing. The subject area is projectile motion, specifically focusing on the horizontal and vertical components of motion.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the equations of motion relevant to the problem, questioning the placement of initial velocities and the calculation of time based on horizontal motion. There is a focus on clarifying the initial vertical velocity and its implications for the height calculation.

Discussion Status

The discussion is active, with participants sharing their calculations and questioning each other's reasoning. Some have provided guidance on how to approach the problem, particularly regarding the correct interpretation of initial velocities. There is no explicit consensus on the final height value, but there is acknowledgment of the calculations presented.

Contextual Notes

Participants are navigating the constraints of the problem, including the assumption of horizontal launch and the effects of gravity on vertical motion. The original poster's confusion regarding the initial vertical velocity is a key point of discussion.

SamL
Messages
11
Reaction score
0

Homework Statement


A ball thrown horizontally at 12.4m/s from the roof of a building lands 20.2m from the base of the building. How tall is the building?

Homework Equations


I think I'm supposed to use these:
horizontal motion: x=x0+vx0t
vertical motion: y=y0=+vy0(t)-.5at^2

The Attempt at a Solution


So this is what I think goes where:
x=20.2
x0= 0
y0= 0
vy0=12.4
a= -9.8
I plugged them in but I'm not sure if they're even in the right spot. After I plug them in I have no idea where to go from there. Please help.
 
Physics news on Phys.org
Horizontal motion is not accelerated or decelerated here. So, if you know the distance horizontally and the velocity horizontally, you can calculate time. From there you can figure out the height

In your solution, you put vy0 as the 12 m/s but it should be vx0
 
I got the time as approx. 1.6290. Is that right? I plugged the time into the vertical equation and got approx. y=-5.8067
 
SamL said:
I got the time as approx. 1.6290. Is that right? I plugged the time into the vertical equation and got approx. y=-5.8067

Right time. How did you manage to get that y value though? Show what you plugged into what.
 
I have y=y0+vyo(t)+.5at^2
Then I plugged in y=0+12.4t+.5(-9.8)t^2
Which then becomes y=12.4t+.5(-9.8)t^2
I then plugged in the time, so y=12.4+.5(-9.8)(1.6290)
 
SamL said:
I have y=y0+vyo(t)+.5at^2
Then I plugged in y=0+12.4t+.5(-9.8)t^2
Which then becomes y=12.4t+.5(-9.8)t^2
I then plugged in the time, so y=12.4+.5(-9.8)(1.6290)

12.4 m/s is vx0. What's vy0? It says 'thrown horizontally'.
 
So it's 0 m/s instead of 12.4?
 
For vy0, yes it is 0 m/s.
 
So it's 0 m/s instead of 12.4 m/s?
 
  • #10
Sorry I didn't mean to post that twice. So I got approx. 13.0034 for the height of the building. Is that correct?
 
  • #11
Looks good ;)
 
  • #12
Thank you guys so much!
 
  • #13
NP. Glad you got it :P
 

Similar threads

Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
15
Views
3K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
8
Views
3K