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

  • Thread starter Thread starter pureouchies4717
  • Start date Start date
  • Tags Tags
    Gravitation
Click For Summary
To calculate the maximum height of a projectile shot upwards at 6000 km/hr, it is first converted to 1666.67 m/s. Using the formula v^2 = 2gy, an initial calculation yields a height of approximately 141.72 km, but the correct answer is 145 km. The discussion highlights the need to consider gravitational acceleration changes with altitude, suggesting that more complex gravitational equations may be necessary. Some participants express uncertainty about their calculations and seek clarification on the correct approach. The conversation emphasizes the importance of accurate gravitational modeling for high-altitude projectile motion.
pureouchies4717
Messages
98
Reaction score
0
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!
 
Physics news on Phys.org
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?

Similar threads

Replies
40
Views
2K
Replies
11
Views
2K
  • · Replies 12 ·
Replies
12
Views
8K
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
1
Views
2K
  • · Replies 12 ·
Replies
12
Views
9K