a project is launched with a velocity of 35m/s at 55° above
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a project is launched with a velocity of 35m/s at 55° above the horizontal what is the maximum height reached by the projectile?
35米每秒的速度和发射角度55度,可以将速度进行矢量分解为重力方向和水平方向,重力方向速度为35×sin55,然后就是向上直抛那个经典算法了,这个你应该会的
再问: 没有水平方向距离我算不出时间。。。然后算不出最高点的距离。。
再答: 不需要水平时间啊,有向上的初速度和重力加速度就可以算出最高点的位置了, 35*sin55=g*t s=0.5*g*t*t就是结果了
再问: 没有水平方向距离我算不出时间。。。然后算不出最高点的距离。。
再答: 不需要水平时间啊,有向上的初速度和重力加速度就可以算出最高点的位置了, 35*sin55=g*t s=0.5*g*t*t就是结果了
a project is launched with a velocity of 35m/s at 55° above
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