Projectile Motion at an angle

  • Thread starter MozAngeles
  • Start date
  • #1
101
0

Homework Statement


When the dried-up seed pod of a scotch broom plant bursts open, it shoots out a seed with an initial velocity of 2.64 m/s at an angle of 30.0˚ below the horizontal. The seed pod is 0.465 m above the ground.
How long does it take for the seed to land?
What horizontal distance does it cover during its flight?


Homework Equations


Voy=Vsin[tex]\theta[/tex]
Vox=Vcos[tex]\theta[/tex]
h = y0 + v0*t + 1/2 * a * t^2
d= Vox*t

The Attempt at a Solution


I used Voy=2.64*sin30˚, Vox= 2.64cos30 I plugged everything into these equations, solved for t using the quadratic equation, and I got t=0.4723 but that was wrong
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
8
Show your calculations.
 
  • #3
101
0
Voy=2.64sin30= 1.33

h = y0 + v0*t + 1/2 * a * t^2
-0.465= 1.33t + 1/2*(-9.8)t^2
-4.9t^2 + 1.33t+ .465=0
t= -1.33-sqrt 1.33^2-(4*-4.9*.465)/ 2*-4.9
t= .04723
 
  • #4
rl.bhat
Homework Helper
4,433
8
Voy=2.64sin30= 1.33

h = y0 + v0*t + 1/2 * a * t^2
-0.465= 1.33t + 1/2*(-9.8)t^2
-4.9t^2 + 1.33t+ .465=0
t= -1.33-sqrt 1.33^2-(4*-4.9*.465)/ 2*-4.9
t= .04723

vo, h and g are in the same direction. So they must have the same sign.
 
  • #5
101
0
Thank you so much, that helped!
 

Related Threads on Projectile Motion at an angle

  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
6
Views
272
Replies
1
Views
1K
Replies
4
Views
3K
  • Last Post
Replies
2
Views
9K
Replies
3
Views
2K
Replies
2
Views
8K
Top