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

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SUMMARY

The maximum height of a projectile shot upwards at a speed of 6000 km/hr is calculated using kinematic equations and gravitational potential energy. The initial speed converts to 1666.67 m/s. Using the formula v² = 2gy, the calculated height is approximately 141.723 km. However, the correct maximum height is 145 km, which accounts for variations in gravitational force with altitude, as indicated by the formula F = GmM/r².

PREREQUISITES
  • Understanding of kinematic equations
  • Knowledge of gravitational potential energy
  • Familiarity with the conversion of units (km/hr to m/s)
  • Basic grasp of gravitational force and its variation with distance
NEXT STEPS
  • Study the derivation and application of kinematic equations in projectile motion
  • Learn about gravitational potential energy and its implications in physics
  • Explore the effects of altitude on gravitational force using F = GmM/r²
  • Practice unit conversions, specifically between km/hr and m/s in physics problems
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Students in physics, educators teaching projectile motion, and anyone interested in understanding the dynamics of objects in motion under gravitational influence.

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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.

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

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 F= { G m M \over r^2 } ? 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 -G m M / r.

Pat
 
thank you guys/girls!
 
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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