Projectile Motion given only x and y distance

Click For Summary
SUMMARY

The discussion centers on solving a projectile motion problem involving Luke Skywalker dropping a bomb from his X-wing. The bomb must reach a point 10m below the opening of a 1m diameter shaft, starting from a height of 200m and a horizontal distance of 500m. The key equations used include the vertical and horizontal motion equations, with constraints on time (greater than 30 seconds), launch angle (greater than 70°), and initial velocity (between 100 and 200 m/s). The challenge lies in solving for the two unknowns, initial velocity (v0) and angle (θ), using the provided equations.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with kinematic equations
  • Basic knowledge of trigonometry
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Study the derivation of projectile motion equations
  • Learn how to apply constraints in kinematic problems
  • Explore numerical methods for solving systems of equations
  • Investigate the effects of varying launch angles on projectile trajectories
USEFUL FOR

Students in physics or engineering courses, educators teaching projectile motion, and anyone interested in applying kinematic principles to real-world scenarios.

tbbtfan14
Messages
1
Reaction score
0

Homework Statement


The question reads: Luke Skywalker is dropping a bomb down a ventilation shaft on the Death Star. The circular shaft has a diameter of 1m. When the bomb is released, Luke's X-wing start-fighter is 200m above the shaft's opening and 500m horizontally from the front edge of the opening. His ship is traveling at a velocity v0 at an angle θ
above the horizontal. If he wants the bomb to reach the front edge of the opening so that the bomb is moving toward a point on the other side of the shaft that is 10m below the opening, find the appropriate values of v0 and θ.


Homework Equations


Velocity in the y direction: vy = 10vx
x= x0 + ((v0)(cosθ))t
y = y0 +((v0)(sinθ))t + .5at^2


Constraints:
time > 30sec
θ > 70°
100 < (v0) < 200 m/s


The Attempt at a Solution


So far I am stuck at: 200 = 4.9t^2 +10((v0)(cosθ))t. I am trying to get an equation to solve for t, but I seem to end up with two variables (v0 and θ) in one equation when I try. Please help this is due tomorrow (9-18) at 10 am!
 
Physics news on Phys.org
Maybe we try reverse position.
The origin of frame of refrence at side of the shaft.

y=uyt -0.5at2
x=uxt => t=x/ux

y=uTanθ- 0.5a(x2/(u2Cos2θ)

2 unknowns and we have 2 points on the path equation.
 
Last edited:

Similar threads

  • · Replies 10 ·
Replies
10
Views
3K
Replies
40
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
7
Views
3K
  • · Replies 29 ·
Replies
29
Views
10K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 13 ·
Replies
13
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K