• Support PF! Buy your school textbooks, materials and every day products Here!

Another projectile motion problem.

  • Thread starter -Dragoon-
  • Start date
  • #1
309
7

Homework Statement


A spring toy is launched from the ground at 2.3 m/s at an angle of 78° to the ground. What is the maximum height reached by the spring toy?

Homework Equations


Vv = vertical velocity, V1 = initial velocity, Dv = vertical distance.
Δt = 2V1Sinθ/(acceleration)
ΔDv = VvΔt + 1/2(acceleration)(Δt^2)
Vv = V1Sinθ

The Attempt at a Solution


First I find the time, since I am given the initial velocity and the launch angle. I get a value of 0.5 seconds. Then I find the vertical velocity by using the equation above, and get about 2.24 m/s. Finally, I use all the information I calculated to find the vertical distance traveled by spring toy, and yield a negative value. What am I doing wrong? I am completely stumped as I haven't been taught any new equations to find the vertical distance. Thanks for all the help!
 

Answers and Replies

  • #2
http://zonalandeducation.com/mstm/physics/mechanics/curvedMotion/projectileMotion/generalSolution/generalSolution.html" [Broken]

The vertical component of the initial velocity is [itex]2.3sin(78^{\circ})[/itex].
[itex]V_{f} = v_{0} + at[/itex]
[itex]t = \frac{V_{f} - v_{0}}{a} = \frac{0 - 2.3sin(78^{\circ})}{-9.81}[/itex].

[itex]D = v_{0}t + \frac{1}{2} at^{2} = (2.3sin(78^{\circ}))(\frac{0 - 2.3sin(78^{\circ})}{-9.81}) + \frac{1}{2}(-9.81)(\frac{0 - 2.3sin(78^{\circ})}{-9.81})^2 \approx .26[/itex] meters.
 
Last edited by a moderator:
  • #3
309
7
http://zonalandeducation.com/mstm/physics/mechanics/curvedMotion/projectileMotion/generalSolution/generalSolution.html" [Broken]

The vertical component of the initial velocity is [itex]2.3sin(78^{\circ})[/itex].
[itex]V_{f} = v_{0} + at[/itex]
[itex]t = \frac{V_{f} - v_{0}}{a} = \frac{0 - 2.3sin(78^{\circ})}{-9.81}[/itex].

[itex]D = v_{0}t + \frac{1}{2} at^{2} = (2.3sin(78^{\circ}))(\frac{0 - 2.3sin(78^{\circ})}{-9.81}) + \frac{1}{2}(-9.81)(\frac{0 - 2.3sin(78^{\circ})}{-9.81})^2 \approx .26 m[/itex]
I used the formula and still get the same negative value.
 
Last edited by a moderator:

Related Threads for: Another projectile motion problem.

  • Last Post
Replies
14
Views
3K
Replies
7
Views
2K
Replies
5
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
6
Views
10K
  • Last Post
Replies
6
Views
5K
Replies
10
Views
1K
Replies
3
Views
1K
Top