Projectile rock hitting a moving van problem

  • Thread starter Thread starter N_L_
  • Start date Start date
  • Tags Tags
    Projectile Rock
Click For Summary
SUMMARY

The discussion centers on a physics problem involving projectile motion, where Suzie throws a rock at a moving police van. The van travels at 3.2 m/s, and the rock is thrown at a speed of 9.0 m/s at a 37-degree angle. The calculations provided indicate that the rock hits the van after 1.25 seconds, with the van's position relative to Suzie being 9.01 meters and the rock's height at impact being 0.7 meters above the ground. The methodology appears sound, but verification of the calculations is necessary.

PREREQUISITES
  • Understanding of projectile motion equations
  • Knowledge of trigonometric functions for resolving vectors
  • Familiarity with kinematic equations
  • Basic concepts of relative motion
NEXT STEPS
  • Review kinematic equations for projectile motion
  • Study vector decomposition in physics
  • Learn about relative motion in two dimensions
  • Practice solving similar projectile motion problems
USEFUL FOR

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

N_L_
Messages
33
Reaction score
0
I have equations for the following problem and an answer, but I am not sure if they are correct:

Suzie (a geology student) is attacking a police van during a demonstration for peace and love at the University. The van is escaping at 3.2 m/s. Suzie throws her rock when the back of the van is 5.0 meters away. The rock leaves her hand at 1.6 meters above the ground, with a speed of 9.0 m/s and an angle of 37 degrees above the horizon. The rock hits and dents the back door of the van. Find the position of the van (relative to Suzie) when the rock hits it. Also find the height of the rock when it hits the van.

Truck:
Xf = Xi + Vo*t + .5 a*t^2
Xf = 5 + 3.2 t + 0

Rock:
Vxf = Vxi = 7.19 m/s after breaking 9.0 m/s at 37 degrees down into x and y components

Xf = Xi + Vxo * t
Xf = 7.19*t

Vyf = Vyo - gt
Vyf = 5.42 - 9.8 t

Yf = Yo + Vyo * t - 4.9 t^2
Yf = 1.6 + 5.42 (1.25) - 4.9 (1.25)^2

The answers I got:

time = 1.25 seconds
Xf relative to Suzie = 9.01 meters
Vf = .7 m above the ground

Is there a mistake?
 
Physics news on Phys.org
I haven't checked the numbers but the method looks ok.

-Dan
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
2K
Replies
11
Views
3K
Replies
15
Views
3K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 8 ·
Replies
8
Views
7K
Replies
6
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K