
#1
Sep2404, 02:07 PM

P: 37

An object is thrown vertically upward such that it has a speed of 58 m/s when it reaches two thirds of it's maximum height above the launch point.
The acceleration of gravity is 9.8 m/s^2 Find the maximum height h. Answer in units of m. I know: V: 58 m/s at y=2/3 H H = ? V = 0 V^2Vo^2 = 2g(1/3H) I know I need to plug in the numbers, but how do I get all of the equation to one side, to find H?? 



#2
Sep2404, 02:24 PM

Mentor
P: 40,878

For example, if you had an equation: 7x^2 = 67bZ, and you wanted to isolate Z, just divide both sides by 67b: (7x^2)/67b = (67bZ)/67b = Z; so Z = (7x^2)/67b. Make sense? 



#3
Sep2404, 02:26 PM

HW Helper
P: 2,280

You got
[tex] V^2 = V_{o}^2 + 2g\frac{2}{3}H [/tex] where V is 58 m/s and [tex] 0 = V_{o}^2 + 2gH [/tex] 



#4
Sep2404, 02:28 PM

HW Helper
P: 2,280

object thrown vertically
Doc Al, how come initial speed = 0 ?




#5
Sep2404, 02:35 PM

Mentor
P: 40,878




Register to reply 
Related Discussions  
Pulling a box vertically  Introductory Physics Homework  2  
Particle projected vertically  Introductory Physics Homework  4  
Net velocity of thrown objectlost  Introductory Physics Homework  10  
Thrown Object  General Physics  6  
Speed of an object thrown down a hole to the center of the earth  Introductory Physics Homework  1 