1* Find the time it takes a particle to reach its greatest height if its speed is 35 m/s and was projected at an angle of 60 degrees.
2* A ball is projected from a point 5 m above horizontal ground with speed 18 m/s at an angle of elevation of 40 degrees. Find the height above the ground the ball is after 2 seconds.
3* A stone is projected horizontally from a point above horizontal ground with speed 32 m/s. The stone takes 2.5 s to reach the ground. Find:
a)the height of the point of projection above the ground,
b)the distance from the point on the ground vertically below the point of projection to the point where the stone reached the ground.
4* A projectile is launched from a point on horizontal ground with speed 150 m/s at an angle of 10 degrees to the horizontal. Find:
a)the time the projectile takes to reach its highest point above the ground,
b)the range of the projectile.
5* A particle is projected fro.m a point O on a horizontal plane with speed 20 m/s at an angle of elevation of 45 degrees. The particle moves freely under gravity until it strikes the ground at a point X. Find:
a)the greatest height above the plane reached by the particle,
b)the distance OX.
幾條好難的英文數學題
2013-08-15 6:27 am
回答 (1)
2013-08-15 8:03 am
✔ 最佳答案
The questions are on the topic of projectile motion in Physics.The horizontal motion is in a uniform velocity.
The vertical motion is in a uniform a = g = -10 m/s²
(Take upward quantity as positive.)
=====
1*
Vertical motion :
u = 35 sin60° m/s, a = -10 m/s², v = 0 m/s, t = ? s
v = u + at
0 = (35 sin60°) + (-10)t
Time taken, t = 3.0 s
=====
2*
Vertical motion :
u = 18 sin40° m/s, a = -10 m/s², t = 2 s, s = ? m
s = ut + (1/2)at²
s = (18 sin40°)(2) + (1/2)(-10)(2)²
s = 3.1 m
Height above the ground after 2 s = 5 + 3.1 = 8.1 m
=====
3*
a)
Vertical motion :
u = 0 m/s, a = -10 m/s², t = 2.5 s, s = ? m
s = ut + (1/2)at²
s = 0 + (1/2)(-10)(2.5)² = -31.3 m
Height = 31.3 m
b)
Horizontal moton :
v = 32 m/s, t = 2.5 s, s = ? m
Distance = vt = (32)(2.5) = 80 m
=====
4*
a)
Vertical motion (to the highest point) :
u = 150 sin10° m/s, v = 0 m/s, a = -10 m/s², t = ? s
v = u + at
0 = 150 sin10° + (-10)t
Time taken, t = 2.6 s
b)
Vertical motion (to hit the ground) :
u = 150 sin10° m/s, s = 0 m, a = -10 m/s², t = ? s
s = ut + (1/2)at²
0 = (150 sin10°)t + (1/2)(-10)(t²)
5t(t - 30 sin10°) = 0
t = 0 (rejected) or t = 30 sin10° = 5.2 s
Horizontal motion until the particle hits the ground :
v = 150 cos10° m/s, t = 5.2 s
Range of the projectile, s = vt = (150 cos10°)(5.2) = 768 m
=====
5*
a)
Vertical motion (to the greatest point) :
u = 20 sin45° m/s, v = 0 m/s, a = -10 m/s², s = ? m
v² = u² + 2as
0 = (20 sin45°)² + 2(-10)s
Greatest height, s = 10 m
b)
Vertical motion (to hit the ground) :
u = 20 sin45° m/s, s = 0 m, a = -10 m/s², t = ? s
s = ut + (1/2)at²
0 = (20 sin45°)t + (1/2)(-10)(t²)
5t(t - 4 sin45°) = 0
t = 0 (rejected) or t = 4 sin45° = 2.8 s
Horizontal motion (to hit the ground) :
v = 20 cos45° m/s, t = 2.8 s
Distance OX, s = vt = (20 cos45°)(2.8) = 40 m
參考: andrew
收錄日期: 2021-04-13 19:37:51
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130814000051KK00384