What is the max height that the cannonball reaches?
What is the total time the cannonball is in the air?
What is the total distance the cannonball travels along the horizontal?
A cannon fires a cannonball at an angle of 45° with an initial velocity of 75 m/s off a cliff that is 150 m high?
2020-08-06 1:36 am
回答 (1)
2020-08-06 1:58 am
1.
Take g = 9.8 m/s², and take all upward quantities to be positive.
Consider the vertical motion (uniform acceleration motion):
Initial velocity, u = 75 sin45° m/s
Final speed, v = 0 m/s
Acceleration, a = -9.8 m/s
v² = u² + 2as
0 = (75 sin45°)² + 2(-9.8)s
s = (75 sin45°)² / [2(9.8)]
s = 143 m
Maximum height = (143 + 150) m = 293 m
====
2.
Consider the vertical motion (uniform acceleration motion):
Initial velocity, u = 75 sin45° m/s
Displacement, s = -150 m
Acceleration, a = -9.8 m/s
s = ut + (1/2)at²
-150 = (75 sin45°)t + (1/2)(-9.8)t²
4.9t² - (75 sin45°)t - 150 = 0
t = {-(-75 sin45°) + √[(-75 sin45°)² - 4(4.9)(-150)]} / [2(4.9)]
Time taken, t = 13 s
====
3.
Horizontal distance = (75 cos45°) (13) = 689 m
Take g = 9.8 m/s², and take all upward quantities to be positive.
Consider the vertical motion (uniform acceleration motion):
Initial velocity, u = 75 sin45° m/s
Final speed, v = 0 m/s
Acceleration, a = -9.8 m/s
v² = u² + 2as
0 = (75 sin45°)² + 2(-9.8)s
s = (75 sin45°)² / [2(9.8)]
s = 143 m
Maximum height = (143 + 150) m = 293 m
====
2.
Consider the vertical motion (uniform acceleration motion):
Initial velocity, u = 75 sin45° m/s
Displacement, s = -150 m
Acceleration, a = -9.8 m/s
s = ut + (1/2)at²
-150 = (75 sin45°)t + (1/2)(-9.8)t²
4.9t² - (75 sin45°)t - 150 = 0
t = {-(-75 sin45°) + √[(-75 sin45°)² - 4(4.9)(-150)]} / [2(4.9)]
Time taken, t = 13 s
====
3.
Horizontal distance = (75 cos45°) (13) = 689 m
收錄日期: 2021-04-24 07:56:52
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200805173613AAHcr9T