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关于物理抛物线的一道物理题,英文的真心求解答aprojectislaunchedinagravitationalfieldgwithaninitialvelocityVoatanangleofAwithrespecttothehorizontal(题读的不是特别明白,是从地面以角度A发射一
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关于物理抛物线的一道物理题,英文的真心求解答
a project is launched in a gravitational field g with an initial velocity Vo at an angle of A with respect to the horizontal(题读的不是特别明白,是从地面以角度A发射一个物体?)
1.write down the equations for displacements along x- and along the y-axis as a function of time t 大概意思就是说把x,y方向的位移方程式写出来.重力是g角度是A
2.大概意思就是说把问题一的式子结合起来,证明物体的移动是抛物线
a project is launched in a gravitational field g with an initial velocity Vo at an angle of A with respect to the horizontal(题读的不是特别明白,是从地面以角度A发射一个物体?)
1.write down the equations for displacements along x- and along the y-axis as a function of time t 大概意思就是说把x,y方向的位移方程式写出来.重力是g角度是A
2.大概意思就是说把问题一的式子结合起来,证明物体的移动是抛物线
▼优质解答
答案和解析
一物体在重力场g中以初始速度V0沿与水平夹角A的方向发射
1.写出沿x与y轴以时间t为参量的位移方程
Vx0 = V0cosA
Vy0 = V0sinA
ax = 0
ay = -g
x = Vx0*t (1)
y = Vy0*t - gt^2/2 (2)
2.
由(1)式得
t = x / Vx0
代入(2)式得
y = Vy0 x / Vx0 - g(x / Vx0)^2/2 = x * tanA - x^2 * g / (2 V0^2 * (cosA)^2)
由此可见物体是沿开口向下的抛物线移动
1.写出沿x与y轴以时间t为参量的位移方程
Vx0 = V0cosA
Vy0 = V0sinA
ax = 0
ay = -g
x = Vx0*t (1)
y = Vy0*t - gt^2/2 (2)
2.
由(1)式得
t = x / Vx0
代入(2)式得
y = Vy0 x / Vx0 - g(x / Vx0)^2/2 = x * tanA - x^2 * g / (2 V0^2 * (cosA)^2)
由此可见物体是沿开口向下的抛物线移动
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