Calculating Time and Velocity in Horizontal Projectile Motion

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SUMMARY

The discussion focuses on calculating time and initial velocity for an object projected horizontally with given displacements. The vertical displacement is 5 cm, and the horizontal displacement is 10 cm, under normal Earth gravity (g = 9.80 m/s²). To find the time (t), the kinematic equation d = vit + 0.5at² is applied, simplifying to t = sqrt(2d/a) for vertical motion. The initial horizontal velocity is determined using the formula v = d/t.

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  • Understanding of kinematic equations
  • Basic knowledge of projectile motion
  • Familiarity with gravitational acceleration (g = 9.80 m/s²)
  • Ability to perform algebraic manipulations
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Awsom Guy
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Hello,
Question:
An object is projected horizontally. After t seconds it has a horizontal displacement of 10cm and a vertical diplacement of 5cm.
a) What is the value of t?
b) What is the magnitude of the initial velocity?

a) my attmept.
:( Nothing Sorry i don't get it.
 
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First off, it can reasonably be assumed that we're talking about an object subject to 1 normal Earth gravity (g)?

If so, this is relatively simple since it's only over a short distance span and you don't have to junk with changing gravity like with high altitude satellite launch problems.
What you know:
a (g=9.80m/s2)
vi (0m/s)
d (5cm down [-5cm])
t (?)
vf (?)

1. Finding time span
The four main kinematic equations use four variables usually with three known.
We know a,vi and d with the intent of finding t, so let's use "d=vit+.5at2"
Since vi equals "0" vertically, vit drops.
Further algebra gets us "t=sqrt(2d/a)"
Crunch your numbers.

2. Finding initial horizontal velocity
Gosh... if you need help here,--- .
(By the way: "d=vt" TF "v=d/t")
 
Last edited by a moderator:
Thanks, but I checked this like a month later, :D
 

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