A strange kind of straight line kinematics with uniform acceleration.

AI Thread Summary
A ball is thrown from a 30-meter tall building and hits the ground after 6 seconds. The problem involves calculating the initial speed and the highest height reached by the ball, considering both upward and downward motion. The relevant kinematic equations are applied, but the user initially struggles with missing variables. After clarification, the correct initial speed is determined to be 25 m/s. The user feels confident to solve the second part of the problem independently.
ZamielTheGrey
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Homework Statement


A ball is thrown upward from a building 30 meters tall and misses the edge on its way down, hitting the ground 6 seconds after it was thrown.

A) With what speed was it thrown?
B) What is its highest height?

Homework Equations


Vf = Vi + at
x = 1/2(Vi + Vf)
x = Vi t + 1/2(at^2)
Vf^2 = Vi^2 + 2ax

Vf = final velocity
Vi = initial velocity
a = acceleration
x = distance
t = time

The Attempt at a Solution


So, it is a two part problem, the upward motion, and then the downward.
I will label the distance from being thrown to the highest point of the arc as x1, which is also the same distance down to being level with where it was thrown from, and then x2 is 30 meters.

Thus goes that to find time, I need to use x = Vi t + 1/2(at^2).
xtotal = x1+30
Vi = 0
t2 = 6 - t1
a = 10 (lets make positive downward)

So here I am stumped by the fact that I do not have three known variable to make use of the equation. How am I supposed to proceed?
 
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Make use of this form of the equation.

x = x0 + v0t + 0.5at2

[edit- you are trying to find v0]
 
Last edited:
Yeah... so we have two missing variable then. An X and Vo :O
 
If the x-axis represents height, you know x0 and you know the final x. v0 is the only unknown. Totally solvable.
 
If I take the base of the skyscraper as the start of the x axis, the ball is thrown up from 30m.

x0 = 30m
xfinal= 0

That completely ignores the up and then down arc of the ball, what am I not understanding?
 
The up and down arc is not relevant. Let the equation do the work.

[edit-- "a = 10 (lets make positive downward)". If +x is pointing up, then a should be negative.]
 
Last edited:
So...

x = x0 + v0t + 0.5at2
-30 = V0 + 0.5*(-10*62)
-30 = V0 - 180
150 = V0

Is this it?
 
Close! You dropped the "t" from v0.
 
Oh, oops.

x = x0 + v0t + 0.5at2
-30 = V0*6 + 0.5*(-10*62)
-30 = V0*6 - 180
150 = V0*6
V0 = 25

Alright, great, thanks. The second part I can easily do on my own. :)
 
  • #10
Good!
 
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