A projectile is shot straight up from the Earth's surfaceby chicagobears34 Tags: earth, projectile, shot, straight, surface 

#1
Apr1613, 12:19 PM

P: 37

1. The problem statement, all variables and given/known data
A projectile is shot straight up from the earth's surface at a speed of 1.10×10^4 km/hr 2. Relevant equations mgh=.5mv^2 3. The attempt at a solution I converted the speed to m/s and 3055.5556 m/s masses cancel out so i get 9.8h=.5v^2 I plugged in v and solved for H and got 476347.95m, which was wrong. I'm not sure if this is how you are supposed to solve this problem, but i can't think of any other ways. 



#2
Apr1613, 12:28 PM

Thanks
P: 5,476

"mgh" works only for low altitudes.




#3
Apr1613, 12:39 PM

P: 37

What other gravitational equations have height in them? 



#4
Apr1613, 12:52 PM

Thanks
P: 5,476

A projectile is shot straight up from the Earth's surface
Newton's law of gravitation?




#5
Apr1613, 01:01 PM

P: 37

Newton's law of gravitation is Fg=GmM/r^2. or Ug= GmM/r
how would I get height out of that? 



#6
Apr1613, 01:03 PM

Thanks
P: 5,476

What is "r"?




#7
Apr1613, 01:11 PM

P: 37

radius and I'm not given any masses. the m would cancel out if I do KE=PE, but the M wouldnt




#8
Apr1613, 01:29 PM

Thanks
P: 5,476

Radius from where?




#9
Apr1613, 01:35 PM

P: 37

r would be radius from center of the earth to the projectile.
so would I do GmM/R=.5mv^2? G=6.673x10^11 M=5.972x10^24 solve for R and subtract by radius of earth? 



#10
Apr1613, 01:42 PM

Thanks
P: 5,476

That would be correct. Note that you could find out what GM is by considering the magnitude of the force of gravity right at the surface of the Earth.




#11
Apr1613, 01:52 PM

P: 37

I converted 1.10x10^4 km/hr to 3055.55555m/s and solved for R and got 85328725.7m and then I subtracted by the Earth Radius which is 6371000m and got 78955725.7 as the final answer but that is still wrong




#12
Apr1613, 02:30 PM

Mentor
P: 11,401

The projectile begins at a certain radius, so it has some initial gravitational potential. It ends up at some new distance with another gravitational potential. The KE from the launch is exchanged for this change in gravitational PE.
So you should write an expression for the change in gravitational PE and relate it to the change in KE. 



#13
Apr1613, 06:04 PM

P: 542

Excuse me for butting in but isn't there a relationship something like E = G(m1)(m2)(1/(r1) 1/(r2))
or you can use calculus to integrate E = Int(G(m1)(m2)/r^2) from the surface of the earth r1 to r2 


Register to reply 
Related Discussions  
Phsyics Kinematics: cannonball shot straight up  Classical Physics  5  
Projectile moving from Earth's surface  Introductory Physics Homework  1  
projectile shot downward  Introductory Physics Homework  3  
Urgent: Projectile fired from Earth's surface  Introductory Physics Homework  18  
A projectile of mass 1.100 kg is shot straight up with an initial speed of 29.0 m/s.  Introductory Physics Homework  3 