Projectiles colliding on a horizontal plane

Click For Summary
SUMMARY

The discussion focuses on a projectile motion problem involving a bullet and a moving target. When the bullet is fired at a speed of 70 m/s at an angle Θ, it must satisfy the condition tanΘ = 4/3 to hit a target moving at 42√2 m/s at a 45-degree angle. The time of collision is calculated to be 10/14 seconds, resulting in a horizontal distance of 30 meters from the firing point O. The correct answers, as per the textbook, are 10/7 seconds for the time and 60 meters for the distance, indicating an error in the initial calculations regarding the target's vertical motion.

PREREQUISITES
  • Understanding of projectile motion equations
  • Knowledge of trigonometric functions (sine, cosine, tangent)
  • Familiarity with the concept of relative velocity
  • Basic algebra for solving equations
NEXT STEPS
  • Review projectile motion equations in detail
  • Study the derivation and application of the tangent function in projectile problems
  • Learn about relative motion in two dimensions
  • Practice similar problems involving collisions of projectiles and moving targets
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and projectile motion, as well as educators looking for examples of real-world applications of these concepts.

Woolyabyss
Messages
142
Reaction score
1

Homework Statement


A bullet is fired from a gun fixed at a point O with speed v m/s at an angle Θ to the horizontal, At the instant of firing , a moving target is 10m vertically above O and traveling with constant speed 42√ 2 m/s at a constant angle 45 degrees to the horizontal.The bullet and target move in the same plane.

(i) if v = 70 m/s, show that if the bullet is to hit the target then tanΘ = 4/3

(ii) Find at what time after firing does the bullet strike the target and calculate the horizontal distance of the bullet from O


Homework Equations



s=ut + (1/2)gt^2


The Attempt at a Solution



If they are to collide then there horizontal velocities must be equal.

70cosΘ = 42√2(cos45) .... cosΘ = 42/70 =3/5

using Pythagoras' sinΘ = 4/5 and tanΘ = 4/3


(ii) There height must be the same when they collide h1 = h2

42√ 2(sin45)t - (1/2)gt^2 + 10 = 70(4/5)t - (1/2)gt^2

(1/2)gt^2 cancel

42t + 10 = 56t ...... t = 10/14 seconds

horizontal distance of the bullet from O 70(3/5)(10/14) = 30m

My book says the answers for (ii) are 10/7 seconds and 60m.

Can anybody spot if I went wrong at any point? Any help would be appreciated.
 
Physics news on Phys.org
Woolyabyss said:
(ii) There height must be the same when they collide h1 = h2

42√ 2(sin45)t - (1/2)gt^2 + 10 = 70(4/5)t - (1/2)gt^2

The target is not in free fall.
 
Thanks I removed that -(1/2)gt^2 and got the right answer.
 

Similar threads

Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 7 ·
Replies
7
Views
5K
Replies
9
Views
5K
Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 15 ·
Replies
15
Views
10K
Replies
2
Views
2K
  • · Replies 29 ·
Replies
29
Views
2K
Replies
6
Views
3K