How Do You Calculate the Impact of a Cannonball on a Moving Car?

  • Thread starter Thread starter jtspenfield
  • Start date Start date
  • Tags Tags
    Cannon Car
Click For Summary
SUMMARY

The discussion focuses on calculating the impact of a cannonball fired at an angle of 38 degrees with an initial velocity of 1000 ft/sec on a moving car traveling at 30 mph. The horizontal distance the cannonball travels to hit the car is determined to be 30,175 ft, taking approximately 38.29 seconds to reach the car. The car, traveling for the same duration, covers a distance of 1,684 ft before being hit. Additionally, the time it takes for the car to move its length of 10 ft is crucial for determining the moment of impact.

PREREQUISITES
  • Understanding of projectile motion equations
  • Knowledge of basic kinematics
  • Familiarity with unit conversions (mph to ft/sec)
  • Ability to apply trigonometric functions in physics
NEXT STEPS
  • Study projectile motion and its equations in detail
  • Learn about the effects of gravity on projectile trajectories
  • Explore simultaneous equations in kinematics
  • Investigate the conversion of units for speed and distance
USEFUL FOR

Students in physics, educators teaching projectile motion, and anyone interested in real-world applications of kinematics and motion analysis.

jtspenfield
Messages
1
Reaction score
0

Homework Statement


A cannon is fired(from pt. A) at an angle of 38deg and with a Vo of 1000 ft/sec. A car leaves (from pt. C) at 30 mph heading towards the cannon. The cannon hits the truck at point B.
a. Determine the horizontal distance R that the cannonball travels to hit the car.
b. How long did it take to hit the car?
c. How far did it travel from point C before it was hit?
d. If length of the car is 10 ft. How long does the car have when it can be hit?

Homework Equations


X= (Vo^2(sin2theta))/g
X= Xo + (Vo)t
X=Xo +volt +1/2(Ac)(t^2)
x=Vt

The Attempt at a Solution


a. Using the first equation, plugging in the initial velocity(1000ft/sec), angle(38) and dividing by gravity(32.152ft/s^2) yields a solution of 30175ft.

b. plugging the distance traveled into the second equation to determine t yields a solution of 38.29s

c. Since the cannon hits the car after 38.29s, then the car must have traveled for that same length of time, initially leaving at 30mph.
Would it be as simple as x=vt to determine the distance? I come up with 1684 ft(after converting from mph to ft/sec) Or is there some way to use simultaneous equations to find Ac, then determine the distance travelled?

d. Not sure how to begin this one, but if the car is 10 ft long, then the time it takes to move 10 ft would seem to be the time when it could be hit.
 
Physics news on Phys.org
jtspenfield said:

Homework Statement


A cannon is fired(from pt. A) at an angle of 38deg and with a Vo of 1000 ft/sec. A car leaves (from pt. C) at 30 mph heading towards the cannon. The cannon hits the truck at point B.
a. Determine the horizontal distance R that the cannonball travels to hit the car.
b. How long did it take to hit the car?
c. How far did it travel from point C before it was hit?
d. If length of the car is 10 ft. How long does the car have when it can be hit?.

Hi jt! Welcome to PF! :smile:

I'm not sure what you've done, but it looks suspicious …

The cannonball's horizontal velocity is constant, and is V0.cos38º.

The cannonball's initial vertical velocity is V0.sin38º, and you have to apply the usual equations to find out the time t when that velocity is reduced to zero.

Then the cannonball returns to the road, and hits the car, at time 2t. :smile:
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
5K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
2
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 9 ·
Replies
9
Views
1K