Finding the max height of a ball launched as a projectile using work-energy

In summary: So, good job!In summary, using the conservation of energy, the maximum height h_max of a ball launched as a projectile with initial speed v at an angle theta above the horizontal can be expressed as h_max = (v^2sin^2(theta))/2g. It is important to remember to square the trig functions and to not break up the initial velocity into its components.
  • #1
ph123
41
0
A ball is launched as a projectile with initial speed v at an angle theta above the horizontal. Using conservation of energy, find the maximum height h_max of the ball's flight.
Express your answer in terms of v, g, and theta.

My energy equation is as follows:

0.5m(v^2)cos(theta) +0.5m(v^2)sin(theta) = 0.5m(v^2)cos(theta) + mgh_max

v^2cos(theta) + v^2sin(theta) = 2gh_max + v^2cos(theta)
h_max = (v^2sin(theta))/2g

This isn't right, somehow. I know that the only velocity at max height is in the x-direction, so max height should be a sine function, right? Did I forget to include something in my energy equation?
 
Physics news on Phys.org
  • #2
ph123 said:
A ball is launched as a projectile with initial speed v at an angle theta above the horizontal. Using conservation of energy, find the maximum height h_max of the ball's flight.
Express your answer in terms of v, g, and theta.

My energy equation is as follows:

0.5m(v^2)cos(theta) +0.5m(v^2)sin(theta) = 0.5m(v^2)cos(theta) + mgh_max

v^2cos(theta) + v^2sin(theta) = 2gh_max + v^2cos(theta)
h_max = (v^2sin(theta))/2g

This isn't right, somehow. I know that the only velocity at max height is in the x-direction, so max height should be a sine function, right? Did I forget to include something in my energy equation?
Looks like you forgot to square the trig functions. At the top of the flight, for example, the KE is
[tex]KE = 1/2m(vcos\theta)^2[/tex]. Also, in the initial case, you don't have to break v into its x and y components. You can just use 1/2mv^2 and get the same result.
 
  • #3
thanks! i got


h_max = (v^2sin^2(theta))/2g

You actually cleared up a big confusion for me in general. I always was using v^2trig(theta) in lots of problems and never understood what was going wrong.
 
  • #4
ph123 said:
thanks! i got


h_max = (v^2sin^2(theta))/2g

You actually cleared up a big confusion for me in general. I always was using v^2trig(theta) in lots of problems and never understood what was going wrong.
Yes, that is correct, and better than my suggestion to not break up v_initial into its components. In which case you would get, not to confuse you,
[tex] 1/2mv^2 = mgh_{max} + 1/2m(vcos\theta)^2 [/tex]
[tex] v^2 = 2gh_{max} + (vcos\theta)^2 [/tex]
[tex] h_{max} = (v^2 - (vcos\theta)^2 )/2g [/tex]
[tex] h_{max} = (v^2(1 - (cos\theta)^2)/2g [/tex]
[tex] h_{max} = (v^2sin^2\theta)/2g [/tex]
which is a bit more tedious than your approach.
 

What is the formula for finding the maximum height of a ball launched as a projectile using work-energy?

The formula for finding the maximum height of a ball launched as a projectile using work-energy is h = (v2sin2θ)/(2g), where h is the maximum height, v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.

How does the work-energy theorem relate to finding the maximum height of a ball launched as a projectile?

The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy. In the case of a ball launched as a projectile, the initial kinetic energy is converted into potential energy at the maximum height. Therefore, by using the work-energy theorem, we can calculate the maximum height of the ball.

What is the significance of the launch angle in determining the maximum height of a ball launched as a projectile?

The launch angle affects the maximum height of the ball because it determines the initial vertical velocity of the ball. The higher the launch angle, the greater the vertical velocity, resulting in a higher maximum height. However, if the launch angle is too high, the ball may not travel as far horizontally.

How does air resistance affect the maximum height of a ball launched as a projectile?

Air resistance can decrease the maximum height of a ball launched as a projectile. As the ball moves through the air, it experiences air resistance, which acts in the opposite direction of its motion. This decreases the ball's kinetic energy, resulting in a lower maximum height.

Can the maximum height of a ball launched as a projectile ever be greater than the initial height?

No, the maximum height of a ball launched as a projectile cannot be greater than the initial height. This is because the ball's initial potential energy is converted into kinetic energy as it is launched, and then back into potential energy at the maximum height. Therefore, the maximum height can only be equal to or less than the initial height.

Similar threads

  • Introductory Physics Homework Help
Replies
13
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
148
  • Introductory Physics Homework Help
Replies
15
Views
20K
  • Introductory Physics Homework Help
2
Replies
39
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
549
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
783
  • Introductory Physics Homework Help
Replies
5
Views
1K
Replies
10
Views
315
  • Introductory Physics Homework Help
2
Replies
53
Views
3K
Back
Top