Kinematics. cannonball was dropped- how far from the wall does it hit ground?

Click For Summary
SUMMARY

The discussion focuses on a physics problem involving projectile motion, specifically a cannonball fired from a castle wall at a speed of 45 m/s and an angle of 32 degrees. The cannonball, when dropped, hits the ground in 1.4 seconds, allowing for calculations of its horizontal distance from the wall and maximum height. The horizontal distance calculated is approximately 185.46 meters, while the maximum height reached by the cannonball is 28.93 meters. These results are derived using kinematic equations for projectile motion.

PREREQUISITES
  • Understanding of kinematic equations for projectile motion
  • Basic trigonometry for resolving velocity components
  • Knowledge of gravitational acceleration (9.8 m/s²)
  • Ability to perform calculations involving time, distance, and velocity
NEXT STEPS
  • Study the derivation and application of kinematic equations in projectile motion
  • Learn how to resolve vectors into horizontal and vertical components
  • Explore the effects of different launch angles on projectile distance and height
  • Investigate real-world applications of projectile motion in engineering and sports
USEFUL FOR

Students studying physics, educators teaching projectile motion concepts, and anyone interested in applying kinematics to real-world scenarios.

aurorabrv
Messages
4
Reaction score
0

Homework Statement


King Arthur's knights fire a cannon from the top of the castle wall. The cannonball is fired at a speed of 45m/s and an angle of 32degrees . A cannonball that was accidentally dropped hits the moat below in 1.4s .

a) How far from the castle wall does the cannonball hit the ground?

b) What is the ball's maximum height above the ground?


Homework Equations



Unsure what to use.

The Attempt at a Solution



xi = ti = 0
vi = 45 m/s
vix = 45cos32 = 38.16
viy = 45sin32 = 23.85
ax = 0
ay = -9.8 m/s
yf = 0
yi = 1/2 (-9.8m/s^2)(1.4s)^2 = 9.6 m

horizontal v 45cos32 = 38.16
initial v 45sin32 = 23.85

time of fall 2.43 seconds

57.86/9.8 = root of 5.9
= 2.43 + 2.43
= 4.86
= 185.4576m


max height
1/2(9.8) x 2.43^2 = 28.93 m
 
Physics news on Phys.org

Similar threads

Replies
12
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
7K
Replies
10
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
22
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
7
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
5
Views
4K