Kinematics - What do these variables mean?

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SUMMARY

The discussion focuses on understanding the kinematic equation x = Xo + v0t + 1/2at², specifically in the context of a projectile motion problem involving a body thrown vertically. The user attempts to calculate the initial velocity (v0) required for a body to pass a height of 9.8 meters twice in 4 seconds, using the acceleration due to gravity (a = 9.8 m/s²). The calculated initial velocity is 22.05 m/s, which differs from the textbook answer of 24 m/s, prompting a discussion on potential errors in the calculation or the textbook itself.

PREREQUISITES
  • Understanding of kinematic equations in physics
  • Basic knowledge of projectile motion
  • Familiarity with the concept of acceleration due to gravity (9.8 m/s²)
  • Ability to solve algebraic equations
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  • Review the derivation of kinematic equations for projectile motion
  • Learn how to apply the quadratic formula in kinematic problems
  • Investigate common mistakes in calculating initial velocity in projectile motion
  • Explore the implications of air resistance on projectile motion calculations
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Students studying physics, educators teaching kinematics, and anyone interested in mastering projectile motion calculations.

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Homework Statement


first off, i don't completely understand what do the variables in the equation do:

x=Xo+volt+1/2at2

the exercise:
A body is thrown up vertically, it passes trough the height 9,8m twice. Time between those two passes is 4 seconds. Whats the starting velocity ?


Homework Equations





The Attempt at a Solution



i presume i have to do it with two equation systems? first off i'd have to calculate the starting velocity needed to get up and down in four seconds. Then i'd calculate the starting speed needed to get to the "seconds starting speed" at 9,8m and that would be the answer.

Bigthnx
 
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x=x_0 + v_0 t + \frac{1}{2} a t^2
x is displacement at time t
x_0 is initial displacement
v_0 is initial velocity
a is acceleration (constant)
t is time

hint: a=9.8 m/s^2 in this case
 
oh.. new to PF.. welcome :smile:
 
great thanks for explaining the equation for me (y) everything seems much clearer now.

So i tried to solve it, but the answer isn't what its supposed to be =/ I mean, it different in the end of the book. Is the book wrong or am i ?

9,8=0 + Vo4 - 1/2*9,8*4^{2}

4Vo = 1/2*9,8*4^{2} + 9,8

Vo= (1/2*9,8*4^{2} + 9,8) / 4 = 22,05 m/s

According to the textbook the answer should be 24 m/s

Thanks again
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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