Question

几条好难的英文数学题

Answer 1

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/铅掘s² (Take upward quantity as positive.) ===== 1* Vertical motion : u = 35 sin60° m/衡毁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/槐拦核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
参考: andrew