Motion equation
posted by Fred .
I assume that is 5.5 m/s, the initial velocity.
Ok, assuming that, then you know the initial KEnergy; 1/2 .62 (5.5)^2 joules.
At the max height, the KE you are given is KE from horizontal motion. So subtract that amount (2.3J) from the initial KE, and you have then the KE that can go into vertical motion. From that, it is easy:
KE available for vertical motion=PE at top, where PEnergy at the top is mass*g*height and you solve for height.
In answering a question such as this one... you have the total PE = 7.1 J ..
Do you then divide this by mass 0.62 kg multiplied by 9.8 ms or by 9.8 squared ??? If it is 9.8 squared why is this so ????
Why does this mass have KE at top ??? should this not be 0 (zero) ???
Just a little confused...
Thanks
Fred
If you know the increase in PE, having derived if from the KE available from vertical motion, then divide PE by m*g to get the height increase. Joules divided by (kg m/s^2) equals meters
Thank you
Respond to this Question
Similar Questions

Motion equation
A stone, which weighs 0.62kg, is thrown, at 5.5ms, its kinetic energy at its maximum height above the surface is 2.3J, what is the height of the stone at this point? 
Physics
Write a projectile motion problem with the initial condition vi=15 m/s at 50° above the horizontal. Solve this problem for ∆x, max height, and time of flight. 
algebra
An object is released into the air at an initial height of 6 feet and an initial velocity of 30 feet per second. The object is caught at a height of 7 feet. Use the vertical motion model, h = 16t^2+vt+s, where h in the height, t is … 
Physics
I have two different questions I can't figure out. 1) Projectile motion: At what initial speed must the basketball player throw the ball at an angle of 55 degrees above the horizontal, to make the foul shot? 
simple harmonic motion.
A 5.80 kg object on a horizontal frictionless surface is attached to a spring with spring constant 860 N/m. The object is displaced from equilibrium 50.0 cm horizontally and given an initial velocity of 9.9 m/s back toward the equilibrium … 
b.c.m school
A 5.0kg object moves on a horizontal frictionless surface under the influence of a spring with force constant 10 3 N/m. The object is displaced 50 cm and given an initial velocity of 10 m/s back toward the equilibrium position. (a) … 
Physics
A. Consider the motion of a projectile. It is fired at t= 0. Its initial speed is 12 m/s and its initial projection angle is 50◦ from the horizontal. The acceleration of gravity is 9.8 m/s^2. What is the maximum height,h, of … 
Physics
A projectile is launched at an angle of 34.0o above the horizontal. The projectile has a mass of 1.50 kg and is given an initial speed of 20.0 m/s. a) What is the initial kinetic energy of the projectile? 
physics
A mass of 3.25 kg is attached to the end of a spring that is stretched 20 cm by a force of 14 N. It is set in motion with an initial position x_0 = 0 and initial velocity (in m/s) v_0 = 11. Find the amplitude, period, and frequency … 
Projectile Motion Help
I was wondering if you could help me solve this projectile motion problem. A projectile is launched with an initial speed of 14.5 m/s at an angle of 35° above the horizontal. The object lands at the same height from which it was launched. …