How Do You Calculate the Maximum Height of a Projectile Shot Upwards?

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Homework Help Overview

The discussion revolves around calculating the maximum height of a projectile shot straight upwards from the Earth's surface at a speed of 6000 km/hr. Participants are exploring the physics of projectile motion and gravitational effects.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss converting units from km/hr to m/s and apply kinematic equations to estimate height. There are questions about the accuracy of their calculations and the assumptions made regarding gravitational acceleration. Some participants express uncertainty about the effects of changing gravitational force with height.

Discussion Status

Several participants have shared their calculations and expressed doubts about their accuracy. There is a recognition of the need to consider gravitational effects more thoroughly, and some guidance has been offered regarding the use of gravitational potential energy in the context of the problem.

Contextual Notes

Participants note the potential for rounding in the provided answer and mention the lack of information regarding gravitational changes at higher altitudes, which may affect their calculations.

pureouchies4717
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A projectile is shot straight up from the Earth's surface at a speed of 6000 km/hr.How high does it go?

i started out by converting km/hr to m/s

so i got 6000km/hr= 1666.666666m/s

then i used this formula:

v^2=2gy

and i got:
y= 141723 m= 141.723km (i just got lucky that my answer was close)

the correct answer wass 145 km

how do you get 145? can someone please help? thanks!
 
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A projectile is shot straight up from the Earth's surface at a speed of 6000 km/hr.How high does it go?

6000km/hr= 1666.666666m/s

So.. there's only 1 direction for velocity, the y-direction.

[tex]y = V_i t + \frac{1/2} a_g t^2[/tex]
and
[tex]V_f = V_i + a_g t[/tex]

So it's a simple kinematics problem with a kick in it if I'm doing it correctly..

So I proceed to find t=170 seconds.. Plug it in and I get 141780 meters or 141.780 km... My god O_o

My problem is that if we were talking about gravitational acceleration past the Earth, we would be given the height but... Well.. yeah O_o

It might be that they rounded... *shrugs* See if anyone else gets a better idea.
 
Last edited:
can anyone else help please? I am positive that i did it wrong and just got lucky
 
nick727kcin said:
can anyone else help please? I am positive that i did it wrong and just got lucky

Have you covered [itex]F= { G m M \over r^2 }[/itex] ? It seems to me that at that speed, you would have to take into account the change o fthe gravitational force with distance.
 
nick727kcin said:
can anyone else help please? I am positive that i did it wrong and just got lucky

I get 145 km using for gravitational potential enegry [itex]-G m M / r[/itex].

Pat
 
thank you guys/girls!
 

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