Projectile Motion: Diver's Leap from a Tower with Vector Components

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Homework Help Overview

The problem involves a diver leaping from a tower with specified initial horizontal and vertical velocities. Participants are tasked with analyzing the diver's motion using vector components after a time interval of one second.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply basic kinematic equations to determine the diver's horizontal and vertical velocity components after one second. Some participants question the appropriateness of the equations used and suggest alternative formulations for calculating the vertical component of velocity.

Discussion Status

Participants are exploring different equations to describe the motion. Some guidance has been provided regarding the use of a specific formula related to acceleration and velocity. There is an ongoing exchange about the validity of the equations and concepts being applied, with no clear consensus reached yet.

Contextual Notes

There is mention of potential confusion regarding the formulas introduced in the discussion, as some participants note they have not encountered certain equations previously. The context suggests a focus on understanding the application of kinematic principles in projectile motion.

ahrog
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Homework Statement


A diver leaps from the top of a tower with an initial horizontal velocity of 5.0 m/s and an upward velocity of 2.0 m/s. Use a vector diagram to indicate the diver's horizontal and vertical velocity components 1.0s after he leaves the tower.

Homework Equations


x=vxt
y= vyt + 1/2gt2
maybe v=d/t

The Attempt at a Solution



I'm not exactly sure how to go about this problem. Do I just use the x and y formula, and that is all?
Vx= x/t
=5/1
=5

Vy= (y-1/2gt2)/t
=(2-1/2(9.8)(1)2)/1
=-2.9

Would those be right?
 
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Not quite right. The best equation to use here is v2 - v1 = a(t2 - t1). So for vx, a =0 and v2 = v1 = 5. (as you said). For vy, a = -9.8 and v2 = 2 - 9.8 = -7.8.
 
Really? They haven't mentioned that formula in what I've done so far...
 
I had never witnessed such a formula either till it clicked - it's just a slight modification to the [tex]v=u+at[/tex] formula.
where:
v = final velocity
u = initial velocity
a = acceleration due to gravity
t = time

And yes, this is the best formula to use in this case. Horizontal velocity will be kept constant at 5ms-1 since we are assuming no resistances, so you already know what that vector is going to be after 1 second. Now you just need to calculate the vertical vector with this equation, which ak1948 already gave as -7.8ms-1.
 

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